Let $n$ be a positive integer, where $n \geq 3$ and let $a_1, a_2, ..., a_n$ be the lengths of sides of some $n$-gon. Prove that $$a_1 + a_2 + ... + a_n \geq \sqrt{2 \cdot (a_1^2 + a_2^2 + ... + a_n^2)} $$
Problem
Source: czechoslovak national mo round
Tags: combinatorics, inequalities
sqing
08.04.2023 16:37
prvocislo wrote: Let $n$ be a positive integer, where $n \geq 3$ and let $a_1, a_2, ..., a_n$ be the lengths of sides of some $n$-gon. Prove that $$a_1 + a_2 + ... + a_n \geq \sqrt{2 \cdot (a_1^2 + a_2^2 + ... + a_n^2)} $$ Czech and Slovak Olympiad 2022/2023
arqady
08.04.2023 19:02
prvocislo wrote: Let $n$ be a positive integer, where $n \geq 3$ and let $a_1, a_2, ..., a_n$ be the lengths of sides of some $n$-gon. Prove that $$a_1 + a_2 + ... + a_n \geq \sqrt{2 \cdot (a_1^2 + a_2^2 + ... + a_n^2)} $$ We need to prove that: $$2\sum_{1\leq i<j\leq n}a_ia_j\geq\sum_{i=1}^na_i^2$$or $$\sum_{i=1}^na_i\left(\sum_{j\neq i}a_j-a_i\right)\geq0.$$
Ritwin
29.03.2024 04:34
Induct downwards on $n$. We may assume $a_1 \leq a_2 \leq \cdots \leq a_n$. Change $a$ to \[ a' = (a_1+a_2, a_3, a_4, \ldots, a_n), \]which when $n \geq 4$ does indeed represent the side lengths of some $(n-1)$-gon, except the special case $a_1=a_2=a_3=a_4$, which can be checked separately. This increases the right side and doesn't change the left side of the desired inequality.
Hence it suffices to prove the $n = 3$ case. Put $(a_1, a_2, a_3) = (x+y, y+z, z+x)$, so \[ 2(x+y+z) \geq 2 \sqrt{x^2+y^2+z^2+xy+yz+zx} \iff xy+yz+zx \geq 0, \]which is true.