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Problem 664

AMC 10/12, early questions

Algebra Difficulty 3.8 Prove it Junior Macedonian Mathematical Olympiad · North Macedonia

Let $a$ , $b$ and $c$ be positive real numbers such that $abc=1$ . Prove that the following inequality holds
$\frac{1}{2}(\sqrt{a} + \sqrt{b} + \sqrt{c}) + \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} \ge 3$
When does equality hold?

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

Since $(1-\sqrt{bc})^2 \ge 0$ it follows that $1+bc \ge 2\sqrt{bc}$ , i.e. $\frac{1}{2\sqrt{bc}} \ge \frac{1}{1+bc}$ . We get that
$\frac{\sqrt{a}}{2} + \frac{1}{1+a} = \frac{1}{2\sqrt{bc}} + \frac{1}{1+a} \ge \frac{1}{1+bc} + \frac{1}{1+a} = \frac{1}{1+\frac{1}{a}} + \frac{1}{1+a} = 1\ldots(1).$
In the same way we prove that
$\frac{\sqrt{b}}{2} + \frac{1}{1+b} \ge 1\ldots(2)$
and
$\frac{\sqrt{c}}{2} + \frac{1}{1+c} \ge 1\ldots(3).$

By adding (1), (2) and (3) we get the required inequality. Let us note that equality holds if and only if $1=\sqrt{bc}$ , i.e. $a=1$ . In the same way we get $b=1$ and $c=1$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.