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For $x_1,x_2,\ldots,x_n$ positive integers, prove that $$\left(\dfrac{x_1^2+x_2^2+\cdots+x_n^2}{x_1+x_2+\cdots+x_n} \right)^{\frac{kn}{t}} \geq x_1x_2\cdots x_n$$ where $k = \max{\{x_1,x_2,\ldots,x_n \}}$ and $t = \min{\{x_1,x_2,\ldots,x_n} \}$. Under which condition does the equality hold?

Attempt

I tried using Cauchy-Schwarz to get $(x_1^2+x_2^2+\cdots+x_n^2)(n) \geq (x_1+x_2+\cdots+x_n)^2$. Thus, $$ \left(\dfrac{x_1^2+x_2^2+\cdots+x_n^2}{x_1+x_2+\cdots+x_n} \right)^{\frac{kn}{t}} \geq\left(\dfrac{(x_1+x_2+\cdots+x_n)^2}{n(x_1+x_2+\cdots+x_n)} \right)^{\frac{kn}{t}} = \left(\dfrac{x_1+x_2+\cdots+x_n}{n}\right )^{\frac{kn}{t}}.$$ Anything else I could try after this?

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1 Answer 1

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By Cauchy-Schwarz inequality, $$\left(x_1^2+x_2^2+\cdots+x_n^2\right)\ge \frac{(x_1+x_2+\cdots+x_n)^2}{n}$$

Therefore $$\left(\dfrac{x_1^2+x_2^2+\cdots+x_n^2}{x_1+x_2+\cdots+x_n} \right)^{\frac{kn}{t}} \geq \left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{\frac{kn}{t}}$$

By AM-GM:

$$\ge (x_1x_2\cdots x_n)^{\frac{k}{t}}$$

Note that $k=\max\{x_1,x_2,\ldots,x_n\}\ge \min\{x_1,x_2,\ldots,x_n\}=t$ with equality if and only if $x_1=x_2=\cdots =x_n$, therefore

$$\ge x_1x_2\cdots x_n$$

The equality holds if and only if $x_1=x_2=\cdots =x_n$.

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