Inequality to prove involving convex notions

**Hugal** · September 8th 2011, 06:16 AM

Hi everyone,

Here's an exercice I have trouble with.
Let $a = (a_1,\dots,a_n)\in\mathbb{R}^n_+$ and define $G(a) = (a_1\dots a_n)^{\frac{1}{n}}$

I have to show that, assuming $a,b\in\mathbb{R}^n_+, G(a+b) \geqslant G(a) + G(b)$ .

I have already tried a looooot of things. I think I'm suppose to use some convex inequalities : my first thought was to take the log of $G(a+b)$ , but it didn't seem to end very well for me.
I've also tried to use other known inequalities, such as $ab \leqslant \displaystyle \frac{a^2+b^2}{2}$ . But I don't think it works.

So know, it's been almost an hour I'm on it, and I would like to have a small hint : I do not want the whole answer but just a tips which can make me progress.

Thank you for your always-so-good answers,

Hugo.

**girdav** · September 8th 2011, 12:07 PM

We have to show that for $0\leq x_j \leq 1$ , we have $\left(\prod_{j=1}^nx_j\right)^{\frac 1n}+\left(\prod_{j=1}^n(1-x_j)\right)^{\frac 1n}\leq 1$ . Take the $n$ -th power, and use $m:=\min_{1\leq j\leq n}x_j$ .

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