1
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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 $x_1=x_2=\cdots =x_n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.