4
$\begingroup$

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

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

  • 0
    It's worth mentioning that this inequality also leads to the matrix inequality $(\det(A+B))^{1/n}\ge(\det A)^{1/n}+(\det B)^{1/n}$ for positive definite matrices $A$ and $B$ of order $n$.2017-03-21

2 Answers 2

4

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.

0

It's just the Holder's inequality: $\prod_{i=1}^n(a_i+b_i)\geq\left(\prod_{i=1}^na_i^{\frac{1}{n}}+\prod_{i=1}^nb_i^{\frac{1}{n}}\right)^n$

  • 0
    @StubbornAtom Holder for two series it's the following: Let a_1>0, b_i>0, \alpha>0 and \beta>0. Prove that: $(a_1+a_2+...+a_n)^{\alpha}(b_1+b_1+...+b_n)^{\beta}\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$. I used Holder for $n$ series: $\{a_1,b_1\}$, $\{a_2,b_2\}$,..., $\{a_n,b_n\}$2017-03-21