Track / Stage 6 / 202 of 400 #1202 of 2000

Problem 1202

National olympiad, first round

Algebra Difficulty 6.3 Prove it Iranian Mathematical Olympiad · Iran

Suppose $a$ , $b$ , $c$ and $d$ are positive real numbers such that $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} + \frac{1}{d+1} = 2$ . Prove that
$\sqrt{\frac{a^2+1}{2}} + \sqrt{\frac{b^2+1}{2}} + \sqrt{\frac{c^2+1}{2}} + \sqrt{\frac{d^2+1}{2}} \geq 3(\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}) - 8.$

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

$\sum \left( \sqrt{\frac{a^2+1}{2}} - \sqrt{a} \right) = \sum \frac{\frac{a^2+1}{2} - a}{\sqrt{\frac{a^2+1}{2}} + \sqrt{a}} = \frac{1}{2} \sum \frac{(a-1)^2}{\sqrt{\frac{a^2+1}{2}} + \sqrt{a}}$
According to Cauchy-Schwarz inequality, $\sqrt{\frac{a^2+1}{2}} + \sqrt{a} \le \sqrt{2(\frac{a^2+1}{2} + a)} = a + 1$ , hence
$\frac{1}{2} \sum \frac{(a-1)^2}{\sqrt{\frac{a^2+1}{2}} + \sqrt{a}} \ge \frac{1}{2} \sum \frac{(a-1)^2}{a+1}.$
On the other hand, $\frac{(a-1)^2}{a+1} = (a-3) + \frac{4}{a+1}$ ; therefore,
$\frac{1}{2} \sum \frac{(a-1)^2}{a+1} = \frac{1}{2} \sum (a-3) + 2 \underbrace{\sum \frac{1}{a+1}}_{2} = \frac{1}{2} (\sum a) - 2.$
Now, it suffices to prove that $\frac{1}{2}(\sum a) - 2 \ge 2(\sum \sqrt{a}) - 8$ , or equivalently, $3 + \frac{1}{4} \sum a \ge \sqrt{a}$ , which can be deduced by the following AM-GM inequality
$3 + \frac{1}{4} \sum a = \underbrace{\sum \frac{a}{a+1}}_{2} + \sum \frac{a+1}{4} = \sum \left( \frac{a}{a+1} + \frac{a+1}{4} \right) \ge \sum \sqrt{a}.$

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