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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 :)

1 Answers 1

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Combining (1) and (2) gives $n^2 \left(a_{1}^{k}+\cdots +a_{n}^k \right) \geq (a_{1}+\cdots +a_{n})^2\cdot\left(a_{1}^{k-2}+\cdots+a_{n}^{k-2}\right).$ Now keep going. In the next step you get $n^3 \left(a_{1}^{k}+\cdots +a_{n}^k \right) \geq (a_{1}+\cdots +a_{n})^3\cdot\left(a_{1}^{k-3}+\cdots+a_{n}^{k-3}\right).$

By induction, you show that for integer powers $p=1,2,\dots,k$ we have $n^p \left(a_{1}^{k}+\cdots +a_{n}^k \right) \geq (a_{1}+\cdots +a_{n})^p\cdot\left(a_{1}^{k-p}+\cdots+a_{n}^{k-p}\right).$

Plugging in $p=k$ gives $n^k \left(a_{1}^{k}+\cdots +a_{n}^k \right) \geq (a_{1}+\cdots +a_{n})^k\cdot\left(a_{1}^{0}+\cdots+a_{n}^{0}\right),$ which is the desired result.

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    @Iuli So really, you already solved it by yourself.2012-09-07