满足两个约束条件的多元不等式

lemondian

设$a_i>0(i=1,2,\cdots ,n,n\geqslant 3,n\inN^*)$满足：$a_1+a_2+\cdots +a_n=(n-1)^2+1$，且$\frac{1}{a_1}+\frac{1}{a_2}+\cdots +\frac{1}{a_n}=2$，求证：
$(n-1)^3+1\leqslant a_1^2+a_2^2+\cdots +a_n^2\leqslant \dfrac{n[(n-1)^2+1]^2}{4(n-1)}$.

Aluminiumor

首先求出 $a_i$ 的范围.
设$$a_1=x, \sum_{i=2}^n{a_i=y. }$$
由$$\sum_{i=2}^n{a_i}\cdot \sum_{i=2}^n{\frac{1}{a_i}}\ge \left( n-1 \right) ^2$$
可知
$$\frac{1}{x}+\frac{\left( n-1 \right) ^2}{y}\le 2$$
$$\Longrightarrow \frac{1}{x}+\frac{\left( n-1 \right) ^2}{\left( n-1 \right) ^2+1-x}\le 2\Longrightarrow \left[ 2x-\left( n^2-2n+2 \right) \right] \cdot \left[ x-1 \right] \le 0$$
$$\Longrightarrow 1\le x\le \frac{n^2-2n+2}{2}$$
易验证两个不等号均可取等.故
$$a_i\in \left[ 1,\frac{n^2-2n+2}{2} \right] $$
$(\mathbb{I})$ 不等式左端：
易知取等条件为 $\left( a_1,a_2,...,a_n \right) =\left( 1,n-1,n-1,...,n-1 \right) $
通过待定系数法, 我们得到
$$a_{i}^{2}\ge \left( 2n-1 \right) a_i+\frac{\left( n-1 \right) ^2}{a_i}-n^2+1 \left( 1\le a_i\le \frac{n^2-2n+2}{2} \right) $$
故
$$\sum_{i=1}^n{a_{i}^{2}}\ge \left( 2n-1 \right) \sum_{i=1}^n{a_i}+\left( n-1 \right) ^2\sum_{i=1}^n{\frac{1}{a_i}}-n\left( n^2-1 \right) =\left( n-1 \right) ^3+1$$
且可以取等.

$(\mathbb{II})$ 不等式右端：
易知取等条件为
$\left( a_1,a_2,...,a_n \right) =\left(\frac{n^2-2n+2}{2},\frac{n^2-2n+2}{2\left( n-1 \right)},\frac{n^2-2n+2}{2\left( n-1 \right)},...,\frac{n^2-2n+2}{2\left( n-1 \right)} \right) $
通过待定系数法, 我们得到
$$a_{i}^{2}\le \frac{n^3-n^2+2}{2\left( n-1 \right)}a_i+\frac{\left( n^2-2n+2 \right) ^3}{8\left( n-1 \right) ^2}\cdot \frac{1}{a_i}-\frac{\left( n^2-2n+2 \right) ^2\left( 2n-1 \right)}{4\left( n-1 \right) ^2}\left( 1\le a_i\le \frac{n^2-2n+2}{2} \right) $$
故
$$\sum_{i=1}^n{a_{i}^{2}}\le \frac{n^3-n^2+2}{2\left( n-1 \right)}\sum_{i=1}^n{a_i}+\frac{\left( n^2-2n+2 \right) ^3}{8\left( n-1 \right) ^2}\sum_{i=1}^n{\frac{1}{a_i}}-\frac{n\left( n^2-2n+2 \right) ^2\left( 2n-1 \right)}{4\left( n-1 \right) ^2}=\frac{n\left[ \left( n-1 \right) ^2+1 \right] ^2}{4\left( n-1 \right)}$$
且可以取等.

大致思路如上，或有疏漏之处，还请见谅.