1995 IMO Problems/Problem 2

Problem

(Nazar Agakhanov, Russia) Let $a, b, c$ be positive real numbers such that $abc = 1$ . Prove that $\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}.$

Solution

Solution 1

We make the substitution $x= 1/a$ , $y=1/b$ , $z=1/c$ . Then $\begin{align*} \frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} &= \frac{x^3}{xyz(1/y+1/z)} + \frac{y^3}{xyz(1/z+1/x)} + \frac{z^3}{xyz(1/x+1/z)} \\ &= \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} . \end{align*}$ Since $(x^2,y^2,z^2)$ and $\bigl( 1/(y+z), 1/(z+x), 1/(x+y) \bigr)$ are similarly sorted sequences, it follows from the Rearrangement Inequality that $\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{1}{2} \left( \frac{y^2+z^2}{y+z} + \frac{z^2+x^2}{z+x} + \frac{x^2+y^2}{x+y} \right) .$ By the Power Mean Inequality, $\frac{y^2+z^2}{y+z} \ge \frac{(y+z)^2}{2(x+y)} = \frac{x+y}{2} .$ Symmetric application of this argument yields $\frac{1}{2}\left( \frac{y^2+z^2}{y+z} + \frac{z^2+x^2}{z+x} + \frac{x^2+y^2}{x+y} \right) \ge \frac{1}{2}(x+y+z) .$ Finally, AM-GM gives us $\frac{1}{2}(x+y+z) \ge \frac{3}{2}xyz = \frac{3}{2},$ as desired. $\blacksquare$

Solution 2

We make the same substitution as in the first solution. We note that in general, $\frac{p}{q+r} = \frac{(p+q+r)}{(p+q+r)-p} - 1 .$ It follows that $(x,y,z)$ and $\bigl(x/(y+z), y/(z+x), z/(x+y)\bigr)$ are similarly sorted sequences. Then by Chebyshev's Inequality, $\frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} \ge \frac{1}{3}(x+y+z) \left(\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} \right) .$ By AM-GM, $\frac{x+y+z}{3} \ge \sqrt[3]{xyz}=1$ , and by Nesbitt's Inequality, $\frac{x}{y+z} + \frac{y}{z+x} + \frac{z}{x+y} \ge \frac{3}{2}.$ The desired conclusion follows. $\blacksquare$