Let $a_i,b_i$ be nonnegative reals. Prove that
$\prod_{i=1}^na_i^{\frac{1}{n}}+\prod_{i=1}^nb_i^{\frac{1}{n}}\le\prod_{i=1}^n(a_i+b_i)^{\frac{1}{n}}$
Let $a_i,b_i$ be nonnegative reals. Prove that
$\prod_{i=1}^na_i^{\frac{1}{n}}+\prod_{i=1}^nb_i^{\frac{1}{n}}\le\prod_{i=1}^n(a_i+b_i)^{\frac{1}{n}}$
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.
It's just the Holder's inequality: $\prod_{i=1}^n(a_i+b_i)\geq\left(\prod_{i=1}^na_i^{\frac{1}{n}}+\prod_{i=1}^nb_i^{\frac{1}{n}}\right)^n$