TITLE: Prove that $\frac{1}{^n\sqrt{1+m}}+\frac{1}{^m\sqrt{1+n}}\geq1$
QUESTION [9 upvotes]: Prove that
$$\dfrac{1}{\sqrt[n]{1+m}}+\dfrac{1}{\sqrt[m]{1+n}}\geq1\quad\forall m,n\in\Bbb N.$$

I've tried to use a common denominator but only got myself into more square problems.

REPLY [6 votes]: In fact,your inequality for any $m,n\in R(>1)$ is hold.
because you can  use Bernoulli inequality
$$(1+x)^a\le 1+ax,0<a<1\Longrightarrow \sqrt[n]{1+m}=(1+m)^{\frac{1}{n}}<1+\frac{m}{n}$$
then
$$\dfrac{1}{\sqrt[n]{m+1}}>\dfrac{1}{1+\frac{m}{n}}=\dfrac{n}{m+n}$$