Track / Stage 6 / 28 of 400 #1028 of 1964

Problem 1028

National olympiad, first round

Algebra Difficulty 6.0 Prove it

Example 1.11.7. Let $a, b, c$ be non-negative real numbers with sum 3. Prove that
$\frac{a+b}{1+b}+\frac{b+c}{1+c}+\frac{c+a}{1+a} \geq 3$

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

SOLUTION. The inequality is equivalent to (homogeneous form)
$\frac{a+b}{b+\frac{a+b+c}{3}}+\frac{b+c}{c+\frac{a+b+c}{3}}+\frac{c+a}{a+\frac{a+b+c}{3}} \geq 3$

Since this problem is cyclic, homogeneous, and holds if $a=b=c$ , we can assume that $c=0$ and $a+b=3$ . In this case, we have to prove that
$\begin{array}{l} \frac{3}{1+b}+b+\frac{a}{1+a} \geq 3 \\ \Leftrightarrow \frac{3}{4-a}+\frac{a}{1+a} \geq a \\ \Leftrightarrow a^{3}-4 a^{2}+3 a+3 \geq 0 \\ \Leftrightarrow a(a-2)^{2}+(3-a) \geq 0 \end{array}$
which is obvious because $a \in[0,3]$ . The equality holds for only one case, $a=b=c=1$ .

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