0

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.

ahmed
  • 1,273
  • 6
  • 10

1 Answers1

4

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

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

user
  • 154,566
  • 2
    This has been asked and answered multiple times. You already answered an identical question: https://math.stackexchange.com/a/2972343/42969. – Martin R Oct 14 '19 at 10:57
  • 1
    @MartinR I had the feeling to have already used that. But I cannot remember exactly when or if it was a similar or an identical question. – user Oct 14 '19 at 11:06
  • 2
    Have you heard about Approach0? This query finds multiple identical questions within seconds. – And don't forget to cast a “close as duplicate” vote on this one! – Martin R Oct 14 '19 at 11:07