generalization of Young inequality

It’s straightforward to extend Young inequality (http://planetmath.org/YoungInequality)to an arbitrary finite number of : provided that $a_{i}>0$ , $c_{i}>0$ and $\\sum_{i=1}^{n}\\frac{1}{c_{i}}=\\frac{1}{r}$ ,

\\left(\\prod_{i=1}^{n}a_{i}\\right)^{r}\\leq r\\sum_{i=1}^{n}\\frac{a_{i}^{c_{i}}}{%c_{i}}

In fact,

$\\displaystyle\\left(\\prod_{i=1}^{n}a_{i}\\right)^{r}$	$\\displaystyle=$	$\\displaystyle\\exp\\left[\\log\\left(\\prod_{i=1}^{n}a_{i}\\right)^{r}\\right]$
	$\\displaystyle=$	$\\displaystyle\\exp\\left[r\\sum_{i=1}^{n}\\log a_{i}\\right]$
	$\\displaystyle=$	$\\displaystyle\\exp\\left[r\\sum_{i=1}^{n}\\frac{1}{c_{i}}\\log\\left(a_{i}^{c_{i}}%\\right)\\right]$
	$\\displaystyle=$	$\\displaystyle\\exp\\left[\\frac{\\sum_{i=1}^{n}\\frac{1}{c_{i}}\\log\\left(a_{i}^{c_{%i}}\\right)}{\\frac{1}{r}}\\right]$
(by Jensen’s inequality and monotonicity of exp)	$\\displaystyle\\leq$	$\\displaystyle\\exp\\left[\\log\\left(\\frac{\\sum_{i=1}^{n}\\frac{1}{c_{i}}a_{i}^{c_{%i}}}{\\frac{1}{r}}\\right)\\right]$
		$\\displaystyle=r\\sum_{i=1}^{n}\\frac{a_{i}^{c_{i}}}{c_{i}}$

Remark: in the case

\\frac{1}{c_{i}}=1\\text{ \\ \\ \\ \\ }\\forall i

one obtains:

\\left(\\prod_{i=1}^{n}a_{i}\\right)^{\\frac{1}{n}}\\leq\\frac{1}{n}\\sum_{i=1}^{n}a_%{i}

that is, the usual arithmetic-geometric mean inequality, which suggestsYoung inequality could be regarded as a generalization of this classical result.Actually, let’s consider the following restatement ofYoung inequality. Having defined: $w_{i}=\\frac{1}{c_{i}}$ , $\\ \\ \\sum_{i=1}^{n}w_{i}=W=\\frac{1}{r}$ , $\\ \\ \\ x_{i}=a_{i}^{\\frac{1}{w_{i}}}$ we have:

\\left(\\prod_{i=1}^{n}x_{i}^{w_{i}}\\right)^{\\frac{1}{W}}\\leq\\frac{1}{W}\\sum_{i=%1}^{n}w_{i}x_{i}

This expression shows that Young inequality is nothing else thangeometric-arithmentic weighted mean (http://planetmath.org/ArithmeticMean) inequality.