Ideas for inequality

Question

Given two non zero natuals $m$ and $n$, it is asked to prove that

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

I don't see how to start. the presence of $k^{\text{th}} $ roots makes it hard to manipulate.

any ideas are welcome.

Answer 1

We have that by Bernoulli inequality in the form $(1+x)^r\le 1+rx$ for $0\le r\le1$

• $\sqrt[n]{m+1}\le 1+\frac m n$
• $\sqrt[m]{n+1}\le 1+\frac n m$

therefore

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