Trouble proving natural number inequality $\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \ge 1$

Question

I came across an inequality and I can't seem to solve it.

For all natural numbers $m, n$,

$$\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \ge 1.$$

I tried isolating roots and then raise both sides to power of m (or n) but that didn't lean anywhere.

Can anyone show me what would be the way to go about this?

Answer 1

Hint: By Bernoulli's inequality, $$\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \geqslant \frac1{1+m/n}+ \frac1{1+n/m} = 1$$