Olympiad Maths Prep

Track / Stage 7 / 236 of 300 #1636 of 2000

Problem 1636

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.6 Prove it

18. If x,y,zx, y, z are all positive real numbers, and x2+y2+z2=1x^{2}+y^{2}+z^{2}=1. Prove:
(a) yzx+xzy+xyz⩾3\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z} \geqslant \sqrt{3};
(b) y2zx2+x2zy2+x2yz2⩾3\frac{y^{2} z}{x^{2}}+\frac{x^{2} z}{y^{2}}+\frac{x^{2} y}{z^{2}} \geqslant \sqrt{3};
(c) y2z3x4+x2z3y4+x2y3z4⩾3\frac{y^{2} z^{3}}{x^{4}}+\frac{x^{2} z^{3}}{y^{4}}+\frac{x^{2} y^{3}}{z^{4}} \geqslant \sqrt{3}

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

18. (a) From (yzx+xzy+xyz)2=∑cyc(xzy−xyz)2+3∑cycx2⩾3\left(\frac{y z}{x}+\frac{x z}{y}+\frac{x y}{z}\right)^{2}=\sum_{\mathrm{cyc}}\left(\frac{x z}{y}-\frac{x y}{z}\right)^{2}+3 \sum_{\mathrm{cyc}} x^{2} \geqslant 3, we know the original inequality holds, with equality if and only if x=y=z=33x=y=z=\frac{\sqrt{3}}{3}.
(b) Squaring the left side of the original inequality, we get
(y2zx2+x2zy2+x2yz2)2=∑cyc(x2zy2−x2yz2)2+3∑cycx3y=∑cyc(x2zy2−x2yz2)2+3∑cyc(x3y−xy)2+2x2−xy)=∑cyc(x2zy2−x2yz2)2+3∑cyc(x3y−xy)2+32∑cyc(x−y)2+3∑cycx2,\begin{aligned} & \left(\frac{y^{2} z}{x^{2}}+\frac{x^{2} z}{y^{2}}+\frac{x^{2} y}{z^{2}}\right)^{2}=\sum_{\mathrm{cyc}}\left(\frac{x^{2} z}{y^{2}}-\frac{x^{2} y}{z^{2}}\right)^{2}+3 \sum_{\mathrm{cyc}} \frac{x^{3}}{y} \\ = & \left.\sum_{\mathrm{cyc}}\left(\frac{x^{2} z}{y^{2}}-\frac{x^{2} y}{z^{2}}\right)^{2}+3 \sum_{\mathrm{cyc}}\left(\sqrt{\frac{x^{3}}{y}}-\sqrt{x y}\right)^{2}+2 x^{2}-x y\right) \\ = & \sum_{\mathrm{cyc}}\left(\frac{x^{2} z}{y^{2}}-\frac{x^{2} y}{z^{2}}\right)^{2}+3 \sum_{\mathrm{cyc}}\left(\sqrt{\frac{x^{3}}{y}}-\sqrt{x y}\right)^{2} \\ & +\frac{3}{2} \sum_{\mathrm{cyc}}(x-y)^{2}+3 \sum_{\mathrm{cyc}} x^{2}, \end{aligned}

From equation (2), we know the original inequality holds, with equality if and only if x=y=z=33x=y=z=\frac{\sqrt{3}}{3}.
(c) Squaring the left side of the original inequality, and performing similar operations as in equation (2), we get
(y2z3x4+x2z3y4+x2y3z4)2⩾3(x5z2y+y5x2z+z5y2x)\left(\frac{y^{2} z^{3}}{x^{4}}+\frac{x^{2} z^{3}}{y^{4}}+\frac{x^{2} y^{3}}{z^{4}}\right)^{2} \geqslant 3\left(\frac{x^{5}}{z^{2} y}+\frac{y^{5}}{x^{2} z}+\frac{z^{5}}{y^{2} x}\right)

From x5z2y+z2+xy⩾3x2\frac{x^{5}}{z^{2} y}+z^{2}+x y \geqslant 3 x^{2}, etc., we have
x5z2y+y5x2z+z5y2x⩾3∑cycx2−∑cycx2−∑cycxy⩾∑cycx2=1\frac{x^{5}}{z^{2} y}+\frac{y^{5}}{x^{2} z}+\frac{z^{5}}{y^{2} x} \geqslant 3 \sum_{\mathrm{cyc}} x^{2}-\sum_{\mathrm{cyc}} x^{2}-\sum_{\mathrm{cyc}} x y \geqslant \sum_{\mathrm{cyc}} x^{2}=1

Combining equations (3) and (4), it is easy to see that the original inequality holds, with equality if and only if x=y=z=33x=y=z=\frac{\sqrt{3}}{3}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.