inequality again.

Posts: 791
Joined: Aug 4, 2011

by hemangsarkar, Sep 5, 2012, 3:09 PM

Q) if, all $x_{i}$ are non zero real numbers,
$\sum_{cyclic}\sqrt{\frac{x_{1}^4}{x_{2}^4 + x_{3}^4 + \dots+ x_{n}^4}} \geq 2n$ .

My solution : by am – gm we can see that –
$\sqrt{\frac{ x_{2}^4 + x_{3}^4 + \dots+ x_{n}^4}{ x_{1}^4}} \le \frac{1}{2} \left(\frac{ x_{2}^4 + x_{3}^4 + \dots+ x_{n}^4}{x_{1}^4} + 1 \right)$ .

So, we must have $\sqrt{\frac{x_{1}^4}{x_{2}^4 + x_{3}^4 + \dots + x_{n}^4}} \geq 2\left(\frac{x_{1}^4}{\sum_{cyclic}x_{i}^4} \right)$ .

Adding cyclically we get,
$\sum_{cyclic}\sqrt{\frac{x_{1}^4}{x_{2}^4 + x_{3}^4 + \dots + x_{n}^4}} \geq 2n$ .