Maths Olympiad Prep

Track / Stage 6 / 282 of 400 #1282 of 1964

Problem 1282

National olympiad, first round
Algebra Difficulty 6.5 Prove it

15. Given that x,y,zx, y, z are positive numbers, and x+y+z=1x+y+z=1, prove that: x4y(1−y2)+y4z(1−z2)+\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+
z4x(1−x2)⩾18\frac{z^{4}}{x\left(1-x^{2}\right)} \geqslant \frac{1}{8}

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

15. By the generalization of Cauchy's inequality, we get
(y+x+z)[(1+y)+(1+z)+(1+x)][(1−y)+(1−z)+(1−x)]⋅[x4y(1−y2)+y4z(1−z2)+z4x(1−x2)]⩾(x+y+z)4\begin{array}{l} (y+x+z)[(1+y)+(1+z)+(1+x)][(1-y)+(1-z)+(1-x)] \cdot \\ {\left[\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+\frac{z^{4}}{x\left(1-x^{2}\right)}\right] \geqslant(x+y+z)^{4}} \end{array}

Since x+y+z=1x+y+z=1, we have
x4y(1−y2)+y4z(1−z2)+z4x(1−x2)⩾18\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+\frac{z^{4}}{x\left(1-x^{2}\right)} \geqslant \frac{1}{8}

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