How prove this inequality $\left(\frac{x^2_{1}+x^2_{2}+\cdots+x^2_{n}}{x_{1}+x_{2}+\cdots+x_{n}}\right)^{x_{1}+x_{2} +\cdots+x_{n}}\ge $

Question

let $x_{1},x_{2},\cdots,x_{n}>0$,show that $$\left(\dfrac{x^2_{1}+x^2_{2}+\cdots+x^2_{n}}{x_{1}+x_{2}+\cdots+x_{n}}\right)^{x_{1}+x_{2} +\cdots+x_{n}}\ge x^{x_{1}}_{1}\cdot x^{x_{2}}_{2}\cdots x^{x_{n}}_{n}$$

My try: $$\Longleftrightarrow (x_{1}+x_{2}+\cdots+x_{n})\ln{\left(\dfrac{x^2_{1}+x^2_{2}+\cdots+x^2_{n}}{x_{1}+x_{2}+\cdots+x_{n}}\right)}\ge \sum_{i=1}^{n}x_{i}\ln{x_{i}}$$

$$(x_{1}+x_{2}+\cdots+x_{n})\ln{(x^2_{1}+x^2_{2}+\cdots+x^2_{n})}-(x_{1}+x_{2}+\cdots+x_{n})\ln{(x_{1}+x_{2}+\cdots+x_{n})}\ge \sum_{i=1}^{n}x_{i}\ln{x_{i}}$$

and I want use Jensen's inequality,But I failed,can you help me? Thank you

Answer 1

Concavity of $\log x$ and weighted Jensen Inequality directly gives: $$ \log\left(\frac{\sum x_i^2}{\sum x_i} \right) \ge \frac{\sum x_i\log x_i}{\sum x_i} \iff \left(\frac{\sum x_i^2}{\sum x_i}\right)^{\sum x_i } \ge \prod x_i^{x_i}$$

Answer 2

Hint: Apply the weighted AM-GM : $a^{a \over a+b} \cdot b^{b \over a+b} \leq a\cdot {a \over a+b} + b\cdot {b \over a+b}$. This proves the case n = 2. Generalize to n.