proving the inequality $\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}}\ge 1$

Question

I wish to prove the following inequality:

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

I tried to use different ideas, I tried using squeeze theorem or assuming $m\gt n$ and working my way from there, but I did not manage to prove this inequality.

Any suggestions?

Answer 1

Well, $(1+m)^{1/n} \le (1+m/n)$, and $(1+n)^{1/m} \le (1+n/m)$. [Do you see why?]

From this the inequality follows:

$$\frac{1}{(1+m)^{1/n}} + \frac{1}{(1+n)^{1/m}} \ge \frac{1}{1+\frac{m}{n}} + \frac{1}{1+\frac{n}{m}} = \frac{n}{n+m} + \frac{m}{n+m} = 1$$

Answer 2

For any $x>-1$ and $s\in[0,1]$, we have $(1+x)^s\leq1+sx$ by Bernoulli's Inequality (with equality cases $s\in\{0,1\}$ and $x=0$). Ergo, if $m$ and $n$ are any real numbers greater than or equal to $1$ (not just positive integers as the notation seems to suggest), then $$\sqrt[m]{1+n}=(1+n)^{\frac1m}\leq 1+\frac{n}{m}$$ and $$\sqrt[n]{1+m}=(1+m)^{\frac1n}\leq 1+\frac{m}{n}\,.$$ The rest is just as the answer by Mike. However, note that the unique equality case is when $(m,n)=(1,1)$.