Track / Stage 6 / 278 of 400 #1278 of 1964

Problem 1278

National olympiad, first round

Algebra Difficulty 6.5 Prove it

20. Let $a, b, c, d$ be non-negative real numbers, satisfying $ab + bc + cd + da = 1$ . Prove:
$\frac{a^{3}}{b+c+d}+\frac{b^{3}}{a+c+d}+\frac{c^{3}}{a+d+b}+\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3} .$

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

20. Since $\frac{a^{3}}{b+c+d}+\frac{b+c+d}{18}+\frac{1}{12} \geqslant 3 \sqrt[3]{\frac{a^{3}}{a+c+d} \cdot \frac{b+c+d}{18} \cdot \frac{1}{12}}=$ $\frac{a}{2}$ , i.e., $\frac{a^{3}}{b+c+d} \geqslant \frac{a}{2}-\frac{b+c+d}{18}-\frac{1}{12}$ , so the left $\geqslant \frac{a+b+c+d}{2}-\frac{1}{18}(3 a+$ $3 b+3 c+3 d)-\frac{4}{12}=\frac{1}{3}(a+b+c+d)-\frac{1}{3}$ . Also, by the assumption, $a b+b c+c d+$ $d a=1$ , i.e., $(a+c)(b+d)=1$ , so, $a+b+c+d=a+c+\frac{1}{a+c} \geqslant 2$ . Therefore, $\frac{a^{3}}{b+c+d}+\frac{b^{3}}{a+c+d}+\frac{c^{3}}{a+b+d}+\frac{d^{3}}{a+b+c} \geqslant \frac{1}{3}$ .

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