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Problem 1008

National olympiad, first round

Algebra Difficulty 6.0 Prove it

Example 19 (24th All-Soviet Olympiad) Let $a_{1}, a_{2}, \cdots, a_{n}$ be positive numbers, and $a_{1}+a_{2}+\cdots+a_{n}=1$ . Prove that: $\frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{2}^{2}}{a_{2}+a_{3}}+\cdots+\frac{a_{n-1}^{2}}{a_{n-1}+a_{n}}+\frac{a_{n}^{2}}{a_{n}+a_{1}} \geqslant \frac{1}{2}$ .

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

Proof that since $a_{1}, a_{2}, \cdots, a_{n}$ are all positive numbers, by the AM-GM inequality we have
$\begin{array}{l} \frac{a_{1}^{2}}{a_{1}+a_{2}}+\frac{a_{1}+a_{2}}{4} \geqslant 2 \sqrt{\frac{a_{1}^{2}}{a_{1}+a_{2}} \cdot \frac{a_{1}+a_{2}}{4}}=a_{1}, \\ \frac{a_{2}^{2}}{a_{2}+a_{3}}+\frac{a_{2}+a_{3}}{4} \geqslant 2 \sqrt{\frac{a_{2}^{2}}{a_{2}+a_{3}} \cdot \frac{a_{2}+a_{3}}{4}}=a_{2}, \\ \cdots \cdots \\ \frac{a_{n}^{2}}{a_{n}+a_{1}}+\frac{a_{n}+a_{1}}{4} \geqslant 2 \sqrt{\frac{a_{n}^{2}}{a_{n}+a_{1}} \cdot \frac{a_{n}+a_{1}}{4}}=a_{n} . \end{array}$
$n$
By adding the above $n$ inequalities in the same direction and noting that $a_{1}+a_{2}+\cdots+a_{n}=1$ , we obtain the desired conclusion.

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