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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 :)

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1 Answer 1

up vote 3 down vote accepted

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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1  
Wait. No link to the Cauchy-Schwarz Master Class? I'm disappointed... :) –  t.b. Sep 7 '12 at 13:41
    
@ByronSchmuland I have one question. I have the felling that what we used is induction. Here is the problem, is induction, or not ? thaks :) –  Iuli Sep 7 '12 at 13:41
    
@t.b. It is on my desk in front of me ... :) –  Byron Schmuland Sep 7 '12 at 13:42
    
@Iuli Yes, a formal proof would use induction on the power. –  Byron Schmuland Sep 7 '12 at 13:42
    
@Iuli So really, you already solved it by yourself. –  Byron Schmuland Sep 7 '12 at 13:45

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