TITLE: Sum of radicals greater than 1
QUESTION [10 upvotes]: Prove that for every $n,m \in \Bbb N $
$$ \frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1 $$

REPLY [3 votes]: For every $a,b \in \mathbb{N}$ we have
$$
\left(1+\frac{b}{a}\right)^a=\sum_{i=0}^a{a\choose i}\left(\frac{b}{a}\right)^i=1+b+\ldots \ge 1+b.
$$
Therefore
$$
(1+b)^{1/a}\le 1+\frac{b}{a}=\frac{a+b}{a} \quad \forall a,b \in \mathbb{N}.
$$
Hence for every $m,n \in \mathbb{N}$ we have
$$
\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \ge \frac{n}{m+n}+\frac{m}{m+n}=1.
$$