I want to show that for each $m,n\in\Bbb{N}$,

$$\large{ \dfrac{1}{\sqrt[n]{1+m}}+\dfrac{1}{\sqrt[m]{1+n}}\geq 1}.$$

I tried induction but it doesn't work. Tried to apply the Bernoulli inequality but it didn't work either. Also tried the AM-GM inequality... Help please.

  • 2
    $\begingroup$ You probably misapplied Bernoulli's inequality. It works here. $\endgroup$ – Daniel Fischer Nov 24 '14 at 12:34
  • $\begingroup$ But how do I make the powers $-1/n, -1/m$ integers so that I can apply the inequality? $\endgroup$ – oyster Nov 24 '14 at 12:38
  • $\begingroup$ $\sqrt[n]{1+m} \le 1+\frac{m}n$ by Bernoulli... $\endgroup$ – Macavity Nov 24 '14 at 12:40

To expand on Daniel Fischer's comment

$$(1 + m)^{1/n} \leq 1 + \frac{m}{n} = \frac{m+n}{n} \ \ \Rightarrow \ \ \frac{1}{(1 + m)^{1/n}} \geq \frac{n}{m+n}$$

Now perform a similar calculation for the other term.

  • $\begingroup$ thanks, I didn't know about the generalized bernoulli inequality... $\endgroup$ – oyster Nov 24 '14 at 12:44

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