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$ a_{1},a_{2},....,a_{n}(n\geq 3) $ are positive numbers that : $(a_{1}+a_{2}+....+a_{n})^2 >\frac{3n-1}{3}(a_{1}^2+a_{2}^2+....+a_{n}^2)$

Prove that for any triple $a_{i},a_{j},a_{k} $ are three edge lengths of some triangle, where natural numbers $ i,j,k $ satifying $ 0< i< j< k\leq n $

It seems a Interesting problem , i tried to prove it in case $n=3$ starting an induction on but i failed at that level

please help me. Thank

Maybe can use Induction

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1 Answer 1

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Since the condition is symmetric about $a_1,\dots,a_n$, it suffices to show that $a_1+a_2>a_3$.

When $n=3$, by Cauchy-Schwarz inequality, $$2\left((a_1+a_2)^2+a_3^2\right)\ge \left((a_1+a_2)+a_3\right)^2 \quad{and}\quad a_1^2+a_2^2\ge \frac{1}{2}(a_1+a_2)^2.$$ Combining the two inequalities above with your condition for $n=3$, we have $(a_1+a_2)^2>a_3^2$, so the conclusion follows.

When $n>3$, note that $\frac{3n-1}{3}=\frac{8}{3}+n-3$ and $$a_1+\dots+a_n=\sqrt{\frac{8}{3}}\times\sqrt{\frac{3}{8}}(a_1+a_2+a_3)+1\times a_4+\dots+1\times a_n.$$ Then by Cauchy-Schwarz inequality, $$\frac{3n-1}{3}\left(\frac{3}{8}(a_1+a_2+a_3)^2+a_4^2+\dots+a_n^2\right)\ge(a_1+\dots+a_n)^2.$$ Combining the inequality above with your condition, we can obtain the condition for $n=3$, which completes the proof.

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