# 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,

 $\displaystyle\left(\prod_{i=1}^{n}a_{i}\right)^{r}$ $\displaystyle=$ $\displaystyle\exp\left[\log\left(\prod_{i=1}^{n}a_{i}\right)^{r}\right]$ $\displaystyle=$ $\displaystyle\exp\left[r\sum_{i=1}^{n}\log a_{i}\right]$ $\displaystyle=$ $\displaystyle\exp\left[r\sum_{i=1}^{n}\frac{1}{c_{i}}\log\left(a_{i}^{c_{i}}% \right)\right]$ $\displaystyle=$ $\displaystyle\exp\left[\frac{\sum_{i=1}^{n}\frac{1}{c_{i}}\log\left(a_{i}^{c_{% i}}\right)}{\frac{1}{r}}\right]$ $\displaystyle\leq$ $\displaystyle\exp\left[\log\left(\frac{\sum_{i=1}^{n}\frac{1}{c_{i}}a_{i}^{c_{% i}}}{\frac{1}{r}}\right)\right]$ $\displaystyle=r\sum_{i=1}^{n}\frac{a_{i}^{c_{i}}}{c_{i}}$

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 suggests Young inequality could be regarded as a generalization of this classical result. Actually, let’s consider the following restatement of Young 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 than geometric-arithmentic weighted mean (http://planetmath.org/ArithmeticMean) inequality.

Title generalization of Young inequality GeneralizationOfYoungInequality 2013-03-22 15:43:08 2013-03-22 15:43:08 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 25 Andrea Ambrosio (7332) Result msc 46E30