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?

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

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

  • $\begingroup$ You might want to justify that inequality, since it is not at all obvious. $\endgroup$ – SBareS Nov 5 '16 at 20:47
  • 1
    $\begingroup$ @SBareS I think it's true since for $x\in[0,1]$ and $n\geq1$, $$(1+n)^x\le1+nx$$ $\endgroup$ – Aforest Nov 6 '16 at 2:57
  • $\begingroup$ Got it, thanks guys $\endgroup$ – Itay4 Nov 6 '16 at 6:45

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