Let $a_i,b_i$ be nonnegative reals. Prove that $\prod_{i=1}^na_i^{1/n}+\prod_{i=1}^nb_i^{1/n}\le\prod_{i=1}^n(a_i+b_i)^{1/n}$

Let $a_i,b_i$ be nonnegative reals. Prove that

$\prod_{i=1}^na_i^{1/n}+\prod_{i=1}^nb_i^{1/n}\le\prod_{i=1}^n(a_i+b_i)^{1/n}$

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Have you tried anything? This looks very straightforward; maybe you should try figuring out what the right hand side equals. – chris Oct 27 '12 at 4:07

We can assume that $a_j,b_j$'s are positive. As $\log$ is concave on $(0,+\infty)$, Jensen's inequality gives $$\tag{*}\frac 1n\sum_{j=1}^n\log c_j\leq \log\left(\frac 1n\sum_{j=1}^nc_j\right),$$ which gives $$\prod_{j=1}^nc_j^{1/n}\leq \frac 1n\sum_{j=1}^nc_j.$$ Applying this to $c_j:=\frac{a_j}{a_j+b_j}$, then to $c_j:=\frac{b_j}{a_j+b_j}$, and adding the inequalities, we get the wanted inequality.

Note that we have equality if and only if we have equality in $(*)$ for the given $c_j$'s.

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