Let $a_{1}, a_{2}, \ldots, a_{n}$ and $k \geq 1$. Prove that (using Chebyshev's inequality): $\large \sqrt[k]{\frac{a_{1}^{k}+a_{2}^{k}+\ldots +a_{n}^{k}}{n}}\geq \frac{a_{1}+a_{2}+\ldots + a_{n}}{n}.$
I think I have a (partial) solution but I don't know if what I obtained really help me. So: $\sqrt[k]{\frac{a_{1}^{k}+a_{2}^{k}+\ldots +a_{n}^{k}}{n}}\geq \frac{a_{1}+a_{2}+\ldots + a_{n}}{n} \Leftrightarrow \frac{a_{1}^{k}+\ldots+a_{n}^{k}}{n} \geq \left(\frac{a_{1}+\ldots +a_{n}}{n}\right)^{k}.$ Now we have to prove that :
$n^{k}\cdot \left(a_{1}^{k}+\ldots+a_{n}^{k}\right) \geq n\cdot \left(a_{1}+\ldots+a_{n}\right)^{k}. $ I want to apply Chebyshev's inequality for: $\displaystyle (a_{1}, \ldots, a_{n})$ and $\displaystyle (a_{1}^{k-1}, \ldots, a_{n}^{k-1})$ $\Rightarrow$
$n \left(a_{1}^{k-1}\cdot a_{1}+\ldots +a_{n}^{k-1} \cdot a_{n} \right) \geq (a_{1}+\ldots +a_{n})\cdot\left(a_{1}^{k-1}+\ldots+a_{n}^{k-1}\right)\tag{1}$
Now I apply Chebyshev's inequality for: $\displaystyle (a_{1}, \ldots, a_{n})$ and $\displaystyle (a_{1}^{k-2}, \ldots, a_{n}^{k-2})$ $\Rightarrow$
$n \left(a_{1}^{k-2}\cdot a_{1}+\ldots +a_{n}^{k-2} \cdot a_{n} \right) \geq (a_{1}+\ldots +a_{n})\cdot\left(a_{1}^{k-2}+\ldots+a_{n}^{k-2}\right)\tag{2}$
$\displaystyle(1)$ can be written like :
$n^{2} \left(a_{1}^{k-1}\cdot a_{1}+\ldots +a_{n}^{k-1} \cdot a_{n} \right) \geq n \cdot (a_{1}+\ldots +a_{n})\cdot\left(a_{1}^{k-1}+\ldots+a_{n}^{k-1}\right)\tag{3}$
But we know that : $n \left(a_{1}^{k-1}+\ldots +a_{n}^{k-1}\right) \geq (a_{1}+\ldots +a_{n})\cdot\left(a_{1}^{k-2}+\ldots+a_{n}^{k-2}\right) \tag{4}$
So, using $\displaystyle(3)$ and $\displaystyle(4)$ we will obtain: $n^{2} \left(a_{1}^{k}+\ldots +a_{n}^{k}\right) \geq n \cdot (a_{1}+\ldots +a_{n})\cdot\left(a_{1}^{k-1}+\ldots+a_{n}^{k-1}\right)\geq \\\left(a_{1}+\ldots+a_{n}\right)^{2}\cdot \left(a_{1}^{k-2}+\ldots+a_{n}^{k-2}\right).$
I think is ok, but what have I do to complete the proof? This proof seems to use induction, or not ? Thanks for your help :)