2 Geometric Inequalities
2.1 Proposals
2.1.1
Proposed by M. S. Klamkin, Avco Research and Development. Amer. Math. Monthly [66: 312].
If \(A\), \(B\), \(C\) are angles of a triangle, show that \[ \csc A/2 + \csc B/2 + \csc C/2 \geq 6. \]
2.1.2
Proposed by W.J. Blundon. Math. Mag. [42: 152–153].
Prove that for every triangle \(ABC\),
\[ \frac{\sin A + \sin B + \sin C}{\sin A \sin B \sin C} \geq 4 \]
2.1.3
Proposed by W. J. Blundon. Amer. Math. Monthly [73: 1122].
Prove that, for every triangle \(ABC\), \(s \leq 2R + (3\sqrt{3} - 4)r\), where \(s\), \(R\), and \(r\) are the semiperimeter, circumradius and inradius, respectively. Equality holds only for the equilateral triangle.
2.1.4
Proposed by L. Carlitz, Duke University. Amer. Math. Monthly [70: 758].
Show that in an acute triangle, \(h_1 + h_2 + h_3 \leq 3(R + r)\), where the \(h_i\) are the altitudes, \(R\) the circumradius, and \(r\) the inradius, and show that the equality sign holds only in the case of an equilateral triangle.
2.1.5
Proposed by G. C. Giri, Midnapore College. Crux Math. [7: 302].
Let \[\begin{align*} a &= \tan \theta + \tan \phi, \\ b &= \sec \theta + \sec \phi, \\ c &= \csc \theta + \csc \phi. \end{align*}\]
If the angles \(\theta\) and \(\phi\) are such that the requisite functions are defined and \(bc \neq 0\), show that \(2a/bc < 1\).
2.1.6
Proposed by Jack Garfunkel. Crux Math. [7: 276].
Let \(m_a\), \(m_b\), \(m_c\) denote the lengths of the medians to sides \(a\), \(b\), \(c\), respectively, of triangle \(ABC\), and let \(M_a\), \(M_b\), \(M_c\) denote the lengths of these medians extended to the circumcircle of the triangle. Prove that \[ M_a / m_a + M_b / m_b + M_c / m_c \geq 4. \]
2.1.7
Proposed by J. L. Brenner. Coll. Math. J. [12: 64].
Assume \(a\), \(b\), \(c > 0\). Prove or disprove that \[ a\cos^2\theta + b\sin^2\theta < c \quad \textrm{implies} \quad \sqrt{a} \cos^2\theta + \sqrt{b} \sin^2\theta < \sqrt{c}. \]
2.1.8
Proposed by K. R. S. Sastry. Coll. Math. J. [14: 260].
Prove that a point \(D \in \overline{BC}\) of triangle \(ABC\) exists so that \(AD\) is the geometric mean between \(BD\) and \(DC\) if and only if \(AC + AB \leq \sqrt{2} BC\). When does equality hold?
2.1.9
Proposed by Jack Garfunkel. Crux Math. [9: 179].
For a triangle \(ABC\) with circumradius \(R\) and inradius \(r\), let \(M = (R - 2r)/2R\). An inequality \(P \geq Q\) involving elements of triangle \(ABC\) will be called strong or weak, respectively, according as \[ P - Q \leq M \] or \[ P - Q \geq M. \]
Prove that the following inequality is strong: \[ \sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2} \geq \frac{3}{4}. \]
Prove that the following inequality is weak: \[ \cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2} \geq \sin B \sin C + \sin C \sin A + \sin A \sin B. \]
2.1.10
Proposed by W. J. Blundon. Amer. Math. Monthly [78: 196].
For any triangle (other than equilateral) with circumcentre \(O\), incentre \(I\), and orthocentre \(H\), let the angles have measures \(\alpha \leq \beta \leq \gamma\). Prove \[ 1 < OH/IO < 3 \textrm{ and } 0 < IH/OH < 2/3 \tag{2.1}\]
\[ \begin{split} 0 < IH/IO < 1 \quad &\textrm{if } \beta < 60 \degree \\ IH = IO \quad &\textrm{if } \beta = 60 \degree \\ 1 < IH/IO < 2 \quad &\textrm{if } \beta > 60 \degree \end{split} \tag{2.2}\] and show that the constant 2 in the last inequality cannot be replaced by a smaller number.
2.1.11
Proposed J. T. Groenman. Crux Math. [8: 209, 278].
\(ABC\) is a triangle with area \(K\) and sides \(a\), \(b\), \(c\) in the usual order. The internal bisectors of angles \(A\), \(B\), \(C\) meet the opposite sides in \(D\), \(E\), \(F\), respectively, and the area of triangle \(DEF\) is \(K^\prime\).
Prove that \[ \frac{3abc}{4(a^3 + b^3 + c^3)} \leq \frac{K^\prime}{K} \leq \frac{1}{4}. \]
If \(a = 5\) and \(3abc/4(a^3 + b^3 + c^3) = 5/24\), determine \(b\) and \(c\), given that they are integers.
2.1.12
Proposed by I. J. Schoenberg, University of Wisconsin. SIAM Review [19: 329].
Let \(P_i = (x_i, y_i)\), \(i = 1, 2, 3\), \(x_1 < x_2 < x_3\), be points in the Cartesian \((x, y)\) plane and let \(R\) denote the radius of the circumcircle \(\Gamma\) of the triangle \(P_1P_2P_3\) (\(R = \infty\) if the triangle is degenerate). Show that \[ \frac{1}{R} < 2 \left|\frac{y_1}{(x_1 - x_2)(x_1 - x_3)} + \frac{y_2}{(x_2 - x_3)(x_2 - x_1)} + \frac{y_3}{(x_3 - x_1)(x_3 - x_2)}\right| \] unless both sides vanish and that \(2\) is the best constant.
2.1.13
Proposed by W. J. Blundon. Crux Math. [10: 195].
The notation being the usual one, prove that each of the following is a necessary and sufficient condition for a triangle to be acute-angled:
\[
IH < r \sqrt{2},
\tag{2.3}\] \[
OH < R,
\tag{2.4}\] \[
\cos^2 A + \cos^2 B + \cos^2 C < 1,
\tag{2.5}\] \[
r^2 + r_a^2 + r_b^2 + r_c^2 < 8R^2
\tag{2.6}\]
\[
m_a^2 + m_b^2 + m_c^2 > 6R^2.
\tag{2.7}\]
2.1.14
Proposed by Jack Garfunkel, Forrest Hills High School. Amer. Math. Monthly [81: 1111].
Let \(a\), \(b\), \(c\) be the sides of a triangle \(ABC\), and let \(m_a\), \(m_b\), \(m_c\) be the medians to sides \(a\), \(b\), \(c\) respectively. Extend the medians so as to meet the circumcircle again, and let these chords be \(M_a\), \(M_b\), \(M_c\) respectively. Show that \[ M_a + M_b + M_c \geq \frac{4}{3} (m_a + m_b + m_c) \] and \[ M_a + M_b + M_c \geq \frac{2}{3} \sqrt{3} (a + b + c). \]
When does equality occur?
2.1.15
Proposed by Vedula N. Murty, Pennsylvania State University. Crux Math. [8: 15].
A triangle has sides \(a\), \(b\), \(c\), semiperimeter \(s\), inradius \(r\), and circumradius \(R\).
Prove that \[ (2a - s)(b - c)^2 + (2b - s)(c - a)^2 + (2c - s)(a - b)^2 \geq 0, \] with equality just when the triangle is equilateral.
Prove that the inequality in (a) is equivalent to each of the following \[ 3(a^3 + b^3 + c^3 + 3abc) \leq 4s(a^2 + b^2 + c^2), \] \[ s^2 \geq 16 Rr - 5r^2. \]
2.1.16
Proposed by L. Carlitz, Duke University. Math. Mag. [36: 264].
Let \(I\) denote the incentre, \(R\) the radius of the circumcircle, and \(r\) the radius of the inscribed circle of the triangle \(ABC\). Show that \(6r \leq AI + BI + CI \leq 3R\). Moreover, equality holds in either place if and only if \(ABC\) is equilateral.
2.1.17
Proposed by W. J. Blundon. Amer. Math. Monthly [73: 1016].
Let \(r\) and \(R\) denote the inradius and circumradius, and define \(k = r/R\), so that \(0 < k < \frac{1}{2}\). Prove that for every triangle \(ABC\), \[ -1 + 4k - k^2 \leq \cos A \cos B + \cos B \cos C + \cos C \cos A \leq k + k^2 \tag{2.8}\] and \[ -1 + 3k - \frac{3}{2} k^2 \leq \cos A \cos B \cos C \leq \frac{1}{2} k^2 \tag{2.9}\] with equality only for equilateral triangles.
2.1.18
Proposed by W. J. Blundon and R. H. Eddy. Nieuw Arch. Wisk. [25: 423].
Let \(g_a\), \(g_b\), \(g_c\) denote the cevians of a triangle \(ABC\) concurrent at the Gergonne point. Prove (in the usual notation) that \[ 8Rr + 11r^2 \leq \sum g_a^2 \leq 4R^2 + 11r^2 \] with equality if and only if the triangle is equilateral.
2.1.19
Proposed by W. J. Blundon and R. H. Eddy. Nieuw Arch. Wisk. [26: 231].
Let \(ABCD\) be a quadrilateral inscribed in a circle of radius \(R\), and circumscribed about a circle of radius \(r\). If \(s\) is the semiperimeter of the quadrilateral, prove the inequalities \[ s \leq \sqrt{4R^2 + r^2} + r \quad \textrm{and} \quad s^2 \geq 8r(\sqrt{4R^2 + r^2} - r) \] and find when equality holds in each case. Hence derive the inequalities \[ s \leq 2R + (4 - 2\sqrt{2}r) \quad \textrm{and} \quad s^2 \geq \frac{32\sqrt{2}}{3} Rr - \frac{16}{3} r^2 \] again stating when equality holds.
2.2 Solutions
2.2.1
Solution by W. J. Blundon. Amer. Math. Monthly [66: 916-917].
In the theorem of Fejes-Tóth [8, p. 11]:
“If \(P\) is an arbitrary point in the plane of a triangle \(ABC\) of inradius \(r\), then \(PA + PB + PC \geq 6r\), with equality only for an equilateral triangle,” we take \(P\) to be the incentre and the result follows at once.
Note: The theorem quoted is a special case of a deeper theorem, [12, p. 104], with the inequality \(PA + PB + PC > 2\sqrt{\sqrt{3}T}\), where \(T\) represents the area of the triangle \(ABC\). This latter theorem is an immediate consequence of the isoperimetric property of the hexagon \(AP_3BP_1CP_2\), where \(P_1\), \(P_2\), \(P_3\) are the images of \(P\) in the sides \(BC\), \(CA\), \(AB\) of the triangle.
2.2.2
Solution by L. Carlitz, Duke University. Math. Mag. [43: 48-49].
Let \(R\) denote the radius of the circumcircle. Then \(a = 2R\sin A\), etc., and the stated inequality becomes \[ (a + b + c)R^2 \geq abc. \tag{2.10}\]
Since \[ a + b + c = 2s, \quad abc = 4R\Delta, \] where \(\Delta\) is the area of \(ABC\), (Equation 2.10) becomes \[ Rs \geq 2\Delta = 2rs, \tag{2.11}\] where \(r\) is the inradius. But (Equation 2.11) is equivalent to the familiar inequality \[ R \geq 2r. \]
2.2.3
Comment by Andrzej Makowski. Amer. Math. Monthly [75: 404].
The result is proved in Blundon [4], Formula 17. Here Blundon shows that the given inequality is the strongest possible linear inequality in \(R\), \(r\) and \(s\).
Summary of solution by W. J. Blundon. Canad. Math. Bull. [8: 615-626].
Working with the variables \(x = R/r\) and \(y = s/r\), the well-known inequality \(s^2 \leq 4R^2 + 4Rr + r^2\) [6, p. 50] becomes \[ y^2 \leq 4x^2 + 4x + 3, \] with equality only for the equilateral triangle; that is, when \(x = 2\). The region defined by the inequality is bounded by the hyperboola \(y^2 - 4(x + 1/2)^2 = 2\). From this equation, it is easy to see that \(y = 2x + 1\) is an asymptote. The ray through \((2, 3\sqrt{3})\) parallel to the asymptote has equation \(y = 2x + (3\sqrt{3} - 4)\). Therefore, we can conclude that for \(x \geq 2\), \(\sqrt{4x^2 + 4x + 3} \leq 2x + (3\sqrt{3} - 4)\). Hence, we obtain the inequality \[ y \leq 2x + (3\sqrt{3} - 4) \iff s \leq 2R + (3\sqrt{3} - 4)r, \] with equality only for the equilateral triangle.
2.2.4
Solution by W. J. Blundon. Amer. Math. Monthly [71: 558].
Let \(H\) be the orthocentre of the triangle \(A_1 A_2 A_3\) and let \(H_1\), \(H_2\), \(H_3\) be the feet of the altitudes through \(A_1\), \(A_2\), \(A_3\) respectively. It is well-known [11, p. 191] that \(\sum A_i H = 2(R + r)\). Since the triangle is acute, \(H\) is an interior point and we may apply the Erdös-Mordell Theorem [6, p. 105] to the point \(H\), obtaining \(\sum A_i H \geq 2\sum HH_i\), with equality only for an equilateral triangle. Thus \[ \sum h_i = \sum A_i H + \sum HH_i \leq (3/2)\sum A_i H = 3(R + r), \] with equality if and only if the triangle is equilateral.
2.2.5
Solution by W. J. Blundon. Crux Math. [8: 318-319].
Let \(u = \tan (\theta /2)\) and \(v \tan (\phi /2)\). It follows from the hypothesis that neither \(u\) nor \(v\) can equal any of \(-1\), \(0\), \(1\). Now \[\begin{align*} a &= \frac{2u}{1 - u^2} + \frac{2v}{1 - v^2} = \frac{2(u + v)(1 - uv)}{(1 - u^2)(1 - v^2)}, \\ b &= \frac{1 + u^2}{1 - u^2} + \frac{1 + v^2}{1 - v^2} = \frac{2(1 + uv)(1 - uv)}{(1 - u^2)(1 - v^2)}, \\ c &= \frac{1 + u^2}{2u} + \frac{1 + v^2}{2v} = \frac{(u + v)(1 + uv)}{2uv}. \end{align*}\]
Therefore, \[ \frac{2a}{bc} = \frac{4uv}{(1 + uv)^2} = 1 - \frac{(1 - uv)^2}{(1 + uv)^2} < 1. \]
The inequality is strict, since \(uv = 1\) implies \(b = 0\), contrary to the hypothesis.
2.2.6
:::
Solution by W. J. Blundon. Crux Math. [8: 307-308].
From \(m_a^2 = (2b^2 + 2c^2 - a^2)/4\) and two similar results, we get \[ a^2 = \frac{8}{9}m_b^2 + \frac{8}{9}m_c^2 - \frac{4}{9}m_a^2, \] we now get \[ M_a m_a = m_a^2 + \frac{a^2}{4} = \frac{8}{9}m_a^2 + \frac{2}{9}m_b^2 + \frac{2}{9}m_c^2, \] and so \[ M_a/m_a = \frac{8}{9} + \frac{2}{9}(m_b^2/m_a^2) + \frac{2}{9}(m_c^2/m_a^2). \]
With this and two similar results, we obtain, with sums cyclic over \(a\), \(b\), \(c\), \[ \sum M_a/m_a = \frac{8}{3} + \frac{2}{9}\sum (m_b^2/m_c^2 + m_c^2/m_b^2) \geq 4, \] since \(x^2 + 1/x^2 \geq 2\), with equality just when \(m_a = m_b = m_c\), that is, just when the triangle is equilateral.
2.2.7
Composite of solutions submitted independently by W. J. Blundon and John P. Mazz (student), Frostburg State College. Coll. Math. J. [13: 211].
Let \(a/c = x^2\), \(b/c = y^2\), where \(x,y > 0\). The inequality \(a\cos^2\theta + b\sin^2\theta < c\) becomes \(x^2cos^2\theta + y^2\sin^2\theta < 1\), and is satisfied by exactly those points \((x,y)\) which are in the first quadrant and interior to the ellipse \(x^2cos^2\theta + y^2\sin^2\theta = 1\). These points are also interior to the triangle formed in the first quadrant by the coordinate axes and the tangent to the ellipse at the point \((1,1)\). This tangent has equation \(xcos^2\theta + y\sin^2\theta = 1\) so \(x\cos^2\theta + y\sin^2\theta < 1\), which is the desired inequality.
2.2.8
Solution by W. J. Blundon. Coll. Math. J. [16: 157].
Each pair \((m,n)\) of positive numbers \(m\) and \(n\) such that \(m + n = 1\) determines a unique point \(D \in \overline{BC}\) such that \(BD = ma\) and \(DC = na\); and, conversely, each point \(D \in \overline{BC}\) determines a unique \((m,n)\) such that \(m > 0\), \(n > 0\), \(m + n = 1\), \(AD = ma\), and \(DC = na\). We regard \((m,n)\) as coordinates of \(D\). By Stewart’s Theorem, \(AD^2\) can be represented, in terms of the coordinates of \(D\), in the form \(mb^2 +nc^2 - mna^2\); and \(AD^2 = BD \cdot DC\) if and only if \(2mna^2 = mb^2 + nc^2\). This equation can be written in the form \[ 2mna^2 = mn(b+c)^2 + (mb-nc)^2. \tag{2.12}\]
It follows that the existence of a point \(D \in \overline{BC}\) such that \(AD^2 = BD \cdot DC\) implies \[ b + c \leq \sqrt{2}a \tag{2.13}\] with equality if and only if \(mb - nc = 0\). Now, the condition for equality is that \(BD : DC = AB : AC\), or, equivalently, equality holds if and only if \(\overline{AD}\) is the internal bisector of angle \(BAC\). If (Equation 2.13) holds, does there exist a point \(D \in \overline{BC}\) such that \(AD^2 = BD \cdot DC\)? The function \(f(m) = b\sqrt{m/(1 - m)} - c\sqrt{(1 - m)/m}\) defined for \(0 < m < 1\) is strictly increasing, as can be seen by differentiation, and all nonnegative numbers are in its range; if (Equation 2.13) holds, then there exists an \(m\) such that \(0 < m < 1\) and \[ 2a^2 - (b + c)^2 = \left(b\sqrt{m/n} - c\sqrt{n/m}\right)^2 \tag{2.14}\] where \(n = 1 - m\). Clearly (Equation 2.14) is equivalent to (Equation 2.12), and so, the answer is in the affirmative. This completes the solution.
2.2.9
Solution by W. J. Blundon. Crux Math. [10: 303-304].
We will use the following well-known relations, where all sums are cyclic over \(A\), \(B\), \(C\): \[ R - 2r \geq 0 \tag{2.15}\] \[ \sum \sin^2\frac{A}{2} = \frac{2R - r}{2R} \tag{2.16}\] \[ \sum \cos^2 \frac{A}{2} = \frac{4R + r}{2R} \tag{2.17}\]
and \(s^2 \leq 4R^2 + 4Rr + 3r^2\) (with equality just when \(R = 2r\)), from which follows \[ \sum \sin B \sin C = \frac{s^2 + 4Rr + r^2}{4R^2} \leq \frac{(R + r)^2}{R^2}. \tag{2.18}\]
For each of the proposed inequalities, it will be seen that equality holds just when \(R = 2r\), that is, just when the triangle is equilateral.
From (Equation 2.15) and (Equation 2.16), \[ P - Q = \sum \sin^2\frac{A}{2} - \frac{3}{4} = \frac{2R - r}{2R} - \frac{3}{4} = \frac{R - 2r}{4R} \leq \frac{R - 2r}{2R} = M \] and the given inequality is strong.
From (Equation 2.15), (Equation 2.17) and (Equation 2.18), \[\begin{align*} P - Q - M &= \sum \cos^2 \frac{A}{2} - \sum \sin B \sin C - M \\ &\geq \frac{4R + r}{2R} - \frac{(R + r)^2}{R^2} - \frac{R - 2r}{2R} \\ &= \frac{(R - 2r)(R + r)}{2R^2} \geq 0 \end{align*}\] and the given inequality is weak.
2.2.10
Solution by Anders Bager. Amer. Math. Monthly [79: 397-398].
We shall use the following relations (for non-equilateral triangles) which are all well-known, or at least easily derivable from known relations.
\[ IO^2 = R^2 - 2Rr \tag{2.19}\]
\[ OH^2 = 9R^2 + 8Rr + 2r^2 - 2s^2 \tag{2.20}\]
\[ IH^2 = 4R^2 + 4Rr + 3r^2 - s^2 \tag{2.21}\]
\[ R > 2r \tag{2.22}\]
\[ 16 Rr - 5r^2 < s^2 < 4R^2 + 4Rr + 3r^2 \textrm{ }[6, p. 50] \tag{2.23}\]
\[ s \textrm{ is} >, =, \textrm{or} < (R + r) \sqrt{3} \textrm{ according as } \beta \textrm{ is} >, =, \textrm{or} < \pi/3. \tag{2.24}\]
(As usual, \(R\), \(r\) and \(s\) denote, respectively, the circumradius, inradius and semiperimeter.)
From (Equation 2.20) and (Equation 2.23) it follows that \[ R^2 - 4r^2 < OH^2 < 9R^2 - 24Rr + 12r^2. \]
Dividing through by \(R^2 - 2Rr > 0\), using (Equation 2.19) and extracting square roots, we have \[ (1+2r/R)^{1/2} < OH/IO < (9-6r/R)^{1/2} \] which implies the first inequality of (Equation 2.1).
The inequality \(IH^2/OH^2 < 4/9\), which is equivalent to the second inequality of (Equation 2.1), is seen to be equivalent to \(s^2 > 4Rr + 19r^2\) by using (Equation 2.20) and (Equation 2.21); this last inequality follows easily from (Equation 2.22) and the first part of (Equation 2.23).
To prove (Equation 2.2) we see that \(IH/OH < 2/3\) implies \[ 3IH < 2OH \leq 2(OH + IH) \quad \textrm{and hence} \quad IH/OH < 2. \]
Now, by (Equation 2.19) and (Equation 2.21) we have that \(IH^2\) is \(<\), \(=\), or \(> IO^2\) according as \(s^2\) is \(>\), \(=\), or \(< 3(R+r)^2\). By (Equation 2.24) it follows that \(IH/OH\) is \(<\), \(=\), or \(> 1\) according as \(\beta\) is \(>\), \(=\), or \(< \pi/3\).
We can show that the constant \(2\) is best possible by considering the triangle with vertices \((0,\epsilon)\), \((1,0)\), and \((-1,0)\); by actual computation \(IH/IO\) can be made arbitrarily close to \(2\) by making \(\epsilon\) sufficiently small.
In Section 14.14 of [6], we find the inequality \(OH \geq IH \sqrt{2}\) with equality if and only if the triangle is equilateral. One is tempted to conclude that \(\sqrt{2}\) is best possible, but it follows from the second inequality of the problem that \(3/2\) is better. This seeming paradox is easily resolved since, when the triangle is equilateral, \(I=H=O\), so that all constants will do.
2.2.11
Solution Crux Math. [9: 249-250].
- by M. S. Klamkin, University of Alberta.
We have \(AF/FB = b/a\) and \(AE/EC = c/a\); hence, with brackets denoting area, \[ \frac{[AFE]}{K} = \frac{AF \cdot AE}{AB \cdot AC} = \frac{bc}{(a + b)(a + c)}. \]
With this and two similar results, we obtain \[ \frac{K^\prime}{K} = 1 - \sum_{\text{cyclic}} \frac{bc}{(a + b)(a + c)} = \frac{2abc}{(b + c)(c + a)(a + b)}, \] and we have to show that \[ \frac{3abc}{4(a^3 + b^3 + c^3)} \leq \frac{2abc}{(b + c)(c + a)(a + b)} \leq \frac{1}{4}. \]
The first inequality will follow from \[ 3(b + c)(c + a)(a + b) \leq 8(a^3 + b^3 + c^3) \tag{2.25}\] and the second from \[ 8abc \leq (b + c)(c + a)(a + b). \tag{2.26}\]
Now (Equation 2.25) and (Equation 2.26) are both known to hold for all nonnegative \(a\), \(b\), \(c\), with equality in each case if and only if \(a = b = c\). According to [6, pp. 12-13], the first is due to A. Padoa (1925) and the second to E. Cesàro (1880).
Proofs for (Equation 2.25) and (Equation 2.26) are not given in [6]. The proof of (Equation 2.26) is nearly trivial (\(2\sqrt{bc} \leq b + c\), etc.), so for completeness we give here only a proof of (Equation 2.25), for which the Padoa reference is not easily accessible.
From the A.M.-G.M. inequality, \[ 6abc \leq 2(a^3 + b^3 + c^3); \tag{2.27}\] and from Problem 6-3, Crux Math [5: 198], \[ 3(b^2c + c^2a + a^2b) \leq 3(a ^3 + b^3 + c^3) \tag{2.28}\] and \[ 3(bc^2 + ca^2 + ab^2) \leq 3(a^3 + b^3 + c^3). \tag{2.29}\]
Now adding (Equation 2.27), (Equation 2.28) and (Equation 2.29) yields (Equation 2.25).
The inequality \(K^\prime/K \leq \frac{1}{4}\) of part (a) was already known. It is included in the following more general result: if \(AD\), \(BE\) and \(CF\) are three concurrent interior cevians of a triangle \(ABC\), then the area of triangle \(DEF\) is a maximum if and only if the point of concurrency is the centroid of \(ABC\). (See the end of solution II of Problem 323, Crux Math. [4: 256], and the end of the solution of Problem 585, Crux Math. [7: 304].) So here we have \(\frac{K^\prime}{K} \leq \frac{1}{4}\) with equality if and only if the incentre coincides with the centroid; that is, if and only if \(ABC\) is equilateral.
- by W. J. Blundon.
The equation we must solve for positive integers \(b\) and \(c\) is equivalent to \[ 18bc = 125 + b^3 + c^3, \] which is itself equivalent to \[ 3bc = (b + c)^2 - 6(b + c) + 36 - \frac{91}{b + c + 6}. \]
Since \(b + c + 6 | 91\), we have \(b + c + 6 = 13\) or \(91\), so \(b + c = 7\) or \(85\), and the corresponding values of \(bc\) are \(12\) and \(2250\). Only the pair \((a + c, bc) = (7, 12)\) is satisfactory, and it leads to the unique solution \(\{b,c\} = \{3,4\}.\)
2.2.12
Solution by W. J. Blundon. SIAM Review [20: 399-400].
First note that \[ |P_1P_2| \geq x_2 - x_1 > 0, \quad |P_2P_3| \geq x_3 - x_2 > 0, \quad |P_3P_1| \geq x_3 - x_1 > 0 \] and strict inequality holds in at least two of these (excluding the straightforward case \(R = \infty\)). The required inequality now follows immediately by means of the well known formulae for the area \(\Delta\) of triangle \(P_1P_2P_3\): \[\begin{align*} 2\Delta &= |y_1(x_3 - x_2) + y_2(x_1 - x_3) + y_3(x_2 - x_1)|, \\ 4\Delta R &= |P_1P_2| \cdot |P_2P_3| \cdot |P_3P_1|. \end{align*}\]
That \(2\) is the best constant follows by considering the special case \(P_1(-2k,0)\), \(P_2(0, 2k^2)\), \(P_3(2k,0)\) where \(k\) is arbitrarily small. Here, \(R = 1 + k^2\) and the right hand side of the proposed inequality reduces to \(1\).
2.2.13
Solution by Walther Janous. Crux Math. [11: 259].
We start with two more necessary and sufficient conditions for a triangle to be acute-angled: \[ \cos A \cos B \cos C > 0, \tag{2.30}\]
\[ a^2 + b^2 + c^2 > 8R^2. \tag{2.31}\]
The validity of (Equation 2.31) is obvious. As for (Equation 2.30), we have (all sums and products are cyclic over \(A\),\(B\),\(C\)): (Equation 2.31) \(\iff 4R^2 \sum \sin^2 A > 8R^2 \iff\) (Equation 2.30).
Proofs of (Equation 2.3) - (Equation 2.7): \[ (a) \iff IH^2 - 2r^2 < 0 \iff (2r^2 - 4r^2 \prod \cos A) - 2r^2 < 0 \iff (f). \]
2.2.14
Solution by W. J. Blundon. Amer. Math. Monthly [83: 59].
Define \(k > 0\) by \(a^2 + b^2 + c^2 = 4k^2\). Then, using \(4m_a^2 = 2b^2 + 2c^2 - a^2\) and two similar formulas, we have \(m_a^2 + m_b^2 + m_c^2 = 3k^2\). The well-known equality inequality \(x^2 + y^2 + z^2 \geq (x + y + z)^2 / 3\) gives \(3k^2 = \sum m_a^2 \geq (\sum m_a)^2 / 3\) and \(4k^2 = \sum a^2 \geq (\sum a)^2 / 3\). Therefore \[ 4k \geq \frac{4}{3} \sum m_a \quad \textrm{and} \quad 4k \geq \frac{2\sqrt{3}}{3} \sum a. \]
Eliminating \(a\) from the equations \(4m_a^2 = 8k^2 - 3a^2\) and \(m_a(M_a - m_a) = (\frac{1}{2} a)^2\), we have \(M_am_a = 2(k^2 + m_a^2) / 3\), whence \[ M_a = \frac{2k}{3} (k/m_a + m_a/k) \geq \frac{4k}{3}. \]
Therefore \[ \sum M_a \geq 4k. \]
Combining (1) and (2) we have the required result. It is clear that equality holds if and only if the triangle is equilateral.
2.2.15
Solution bt W. J. Blundon. Crux Math. [9: 49-50].
We will show that the three inequalities \[ (2a - s)(b - c)^2 + (2b - s)(c - a)^2 + (2c - s)(a - b)^2 \geq 0 \] \[ 4s(a^2 + b^2 + c^2) \geq 3(a^3 + b^3 + c^3 + 3abc) \] \[ s^2 \geq 16Rr - 5r^2 \] are equivalent, and then prove that one of them is (and hence all three are) valid, with equality if and only if the triangle is equilateral. In establishing equivalence, we will use the well-known relations \[ a + b + c = 2s, \quad bc + ca + ab = s^2 + 4Rr + r^2, \quad abc = 4Rrs, \] from which follow \[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(bc + ca + ab) = 2(s^2 - 4Rr - r^2) \] and \[\begin{align*} a^3 + b^3 + c^3 &= (a + b + c)^3 - 3(a + b + c)(bc + ca + ab) + 3abc \\ &= 2(s^3 - 6Rrs - 3r^2s). \end{align*}\]
Now (1) \(\iff\) (2) follows from (4) and, since \(s > 0\), (2) \(\iff\) (3) follows from (5).
At this point to establish the validity of (1)-(3), we could simply refer to [6, p. 50] where a proof of (3) is given. We will instead establish (1) directly, because it will lead to an interesting generalization. For this purpose, we will use Schur’s Inequality, [9] or [14]: If t is any real number and \(x,y,z > 0\), then \[ x^t (x - y)(x - z) + y^t (y - z)(y - x) + z^t (z - x)(z - y) \geq 0, \] with equality if and only if \(x = y = z\).
Let \(x = s - a\), \(y = s - b\), \(z = s - c\); then \(x,y,z > 0\) and \(a = y + z\), \(b = z + x\), \(c = x + y\). With cyclic sums again, we have \[\begin{align*} \sum (2a - s)(b - c)^2 &= \sum (y + z - x)(z - y)^2 \\ &= \sum (y^3 + z^3 + 2xyz - y^2z - yz^2 - xy^2 - z^2x) \\ &= 2 \sum x^3 - 2 \sum yz(y + z) + 6xyz \\ &= 2 \sum x(x - y)(x - z), \end{align*}\] and (1) follows from (6) with \(t = 1\).
More generally, for any real number \(t\) we get from (6) \[ \sum (s - a)^t (b - a)(c - a) \geq 0, \] with equality if and only if \(a = b = c\), and this is equivalent to (1)-(3) when \(t = 1\).
2.2.16
Solution by W. J. Blundon. Math. Mag. [37: 121].
To prove the first inequality, we apply the Erdös-Mordell inequality [6, p. 105] to the incentre \(I\) of the triangle \(A_1A_2A_3\) to obtain \[ IA_1 + IA_2 + IA_3 \geq 2(r + r + r) = 6r. \]
Since the orthocentre \(H\) of the triangle \(A_1A_2A_3\) is the incentre of the pedal triangle \(H_1H_2H_3\) and since the circumradius of this triangle is \(\frac{1}{2} R\), the second inequality is equivalent to \(HH_1 + HH_2 + HH_3 \leq \frac{3}{2} R\). To prove this we use the euler inequality \(R \geq 2r\) and also the relation \(HA_1 + HA_2 + HA_3 = 2r + 2R\). Applying the Erdös-Mordell inequality to the point \(H\) of the triangle \(H_1H_2H_3\), we have \[ HH_1 + HH_2 + HH_3 \leq \frac{1}{2} (HA_1 + HA_2 + HA_3) = r + R \leq \frac{3}{2} R. \]
2.2.17
Solution by M. G. Greening, University of New South Wales. Amer. Math. Monthly [75: 196-197].
Let \(I\), \(G\), \(H\) represent the incentre, centroid, and orthocentre, respectively. Then, \[ 4R^2\cos A \cos B \cos C = 2r^2 - IH^2 = 9IG^2 + 12Rr - 4R^2 - 6r^2, \] giving the inequality (2). (See editorial note.) As both \(IH = 0\) and \(IG = 0\) if and only if the triangle is equilateral, the full result follows.
For (1), we have \[\begin{align*} 2\cos A \cos B \cos C &= 1 - \cos^2 A - \cos^2 B - \cos^2 C \\ &= 2 \sum \cos A \cos B - (\sum \cos A)^2 + 1 \\ &= 2 \sum \cos A \cos B - (k + 1)^2 + 1, \end{align*}\] giving \(2 \sum \cos A \cos B = 2 \cos A \cos B \cos C + k^2 + 2k\). Substitution in (2) then gives (1) with the condition for equality preserved.
Editorial Note. The following result \[ IH^2 = IA^2 + AH^2 - 2IA \cdot AH \cos I AH \] is given in [16, p. 251]. Since this book is probably not accessible to most readers, we sketch a proof. Now \[ IA = \frac{r}{\sin \frac{1}{2} A} = 4R\sin \frac{1}{2}B \sin \frac{1}{2}C, \space AH = 2R\cos A, \] and \[ \angle IAH = \frac{1}{2} A - (90 \degree - C) = \frac{1}{2} A - \frac{1}{2} (A + B + C) + C = \frac{1}{2} (C - B), \] therefore, \[ IH^2 = 16R^2 \sin^2 \frac{1}{2} B \sin^2 \frac{1}{2} C + 4R^2 \cos^2 A \] \[ - 16R^2 \sin \frac{1}{2} B \sin \frac{1}{2} C \cos A \cos \frac{1}{2} (C - B) \] which after some simplification, reduces to \[ IH^2 = 32R^2 \sin^2 \frac{1}{2} A \sin^2 \frac{1}{2} B \sin^2 \frac{1}{2} C - 4R^2 \cos A \cos B \cos C. \]
But \(\sin \frac{1}{2} A \sin \frac{1}{2} B \sin \frac{1}{2} C = \frac{r}{4R}\) [4, p. 21], hence \[ IH^2 = 2r^2 - 4R^2\cos A \cos B \cos C \] as required.
The right-hand side of inequality (2) is given in [6, p. 25].
2.2.18
Solution by W. J. Blundon and R. H. Eddy. Nieuw Arch. Wisk. [26: 354-355].
Using the formula (proved below) \[ \sum g_a^2 = s^2 - 8Rr + 2r^2 + \frac{r}{R}(s^2 + r^2) \] and the well-known inequality \(s^2 \leq 4R^2 + 4Rr + 3r^2\), we have \[ \sum g_a^2 \leq 4R^2 + 9r^2 + \frac{4r^3}{R} \] so that \[ (4R^2 + 11r^2) - \sum g_a^2 \geq \frac{2r^2}{R}(R - 2r) \geq 0, \] with equality only for the equilateral triangle. Again, using the formula \(s^2 \geq 16Rr - 5r^2\), we have \[ \sum g_a^2 \geq 8Rr + 13r^2 - \frac{4r^3}{R}, \] so that \[ \sum g_a^2 - (8Rr + 11r^2) \geq \frac{2r}{R}(R - 2r) \geq 0, \] with equality only for the equilateral triangle. This completes the proof.
Proof of the formula.
It follows from the law of cosines that \[ g_a^2 = (s - b)^2 + c^2 - 2c(s - b)(a^2 + c^2 - b^2)/2ac. \]
Putting \(2(s - b) = (a + c - b)\), we obtain \[ 4g_a^2 = -a^2 + 3b^2 + 3c^2 - 2bc + 2(b^3c^2 + b^2c^3 - b^4c - bc^4)/abc. \]
Thus, \[ 4\sum g_a^2 = 5\sum a^2 - 2\sum ab + 2(\sum a^3b^2 - \sum a^4b)/abc. \]
Put \(t_1 = a + b + c = 2x\), \(t_2 = ab + bc + ca = s^2 + 4Rr + r^2\), \(t_3 = 4Rrs\).
Then \(\sum a^2 = t_1^2 - 2t_2 = 2s^2 - 8Rr - 2r^2\). Also, \[\begin{align*} \sum a^3b^3 - \sum a^4b \\ &= 4t_1t_2^2 - 6t_2t_3 - t_1^2t_3 - t_1^3t_2 \\ &= -4Rrs^3 + 8R^2s^3 + 32R^2r^2s + 40Rr^3s + 8r^4s. \end{align*}\]
Hence, after some simplification, we obtain \[ \sum g_a^2 = s^2 - 8Rr + 2r^2 + \frac{r}{R}(s^2 + r^2). \]
2.2.19
Proof of the formula.
It follows from the law of cosines that \[ g_a^2 = (s - b)^2 + c^2 - 2c(s - b)(a^2 + c^2 - b^2)/2ac. \]
Putting \(2(s - b) = (a + c - b)\), we obtain \[ 4g_a^2 = -a^2 + 3b^2 + 3c^2 - 2bc + 2(b^3c^2 + b^2c^3 - b^4c - bc^4)/abc. \]
Thus, \[ 4\sum g_a^2 = 5\sum a^2 - 2\sum ab + 2(\sum a^3b^2 - \sum a^4b)/abc. \]
Put \(t_1 = a + b + c = 2x\), \(t_2 = ab + bc + ca = s^2 + 4Rr + r^2\), \(t_3 = 4Rrs\).
Then \(\sum a^2 = t_1^2 - 2t_2 = 2s^2 - 8Rr - 2r^2\). Also, \[\begin{align*} \sum a^3b^3 - \sum a^4b \\ &= 4t_1t_2^2 - 6t_2t_3 - t_1^2t_3 - t_1^3t_2 \\ &= -4Rrs^3 + 8R^2s^3 + 32R^2r^2s + 40Rr^3s + 8r^4s. \end{align*}\]
Hence, after some simplification, we obtain \[ \sum g_a^2 = s^2 - 8Rr + 2r^2 + \frac{r}{R}(s^2 + r^2). \]