# Help proving an inequality

I need to prove this inequality:

$$\bigg(\frac{1}{1+n}\bigg)^{\frac{1}{m}}+\bigg(\frac{1}{1+m}\bigg)^{\frac{1}{n}} \geq 1$$

I tried ^${nm}$ then make the LCD and because $n$ and $m$ and natural, then it must be positive so it equals to it's absolute value and finally use the triangle inequality.

Any ideas?

• Please indicate what restrictions there are on $n$ and $m$. If $n=m=\frac12$ then the left hand side becomes $$2\times\left(\frac{1}{1+\frac12}\right)^2=2\times\left(\frac{2}{3}\right)^2=2\times\frac49=\frac89$$ – Ian Miller Nov 5 '16 at 15:38
• n and m are natural numbers – Itay4 Nov 5 '16 at 15:43

$$\bigg(\frac{1}{1+n}\bigg)^{\frac{1}{m}}+\bigg(\frac{1}{1+m}\bigg)^{\frac{1}{n}} \geq \frac{1}{1+\frac{n}{m}}+\frac{1}{1+\frac{m}{n}}=1$$
• @SBareS I think it's true since for $x\in[0,1]$ and $n\geq1$, $$(1+n)^x\le1+nx$$ – Aforest Nov 6 '16 at 2:57