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I came across an inequality and I can't seem to solve it.

For all natural numbers $m, n$,

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

I tried isolating roots and then raise both sides to power of m (or n) but that didn't lean anywhere.

Can anyone show me what would be the way to go about this?

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  • $\begingroup$ For all natural numbers $m$, $n$? $\endgroup$
    – Caleb Stanford
    Nov 4, 2016 at 15:07
  • $\begingroup$ yes. I forgot to add that $\endgroup$
    – Joe Shmoe
    Nov 4, 2016 at 15:08
  • $\begingroup$ @6005 isn't there a $\forall$? $\endgroup$
    – Joe Shmoe
    Nov 4, 2016 at 15:10
  • $\begingroup$ Yes, there is. However, when typing math it is almost always clearer to write it out in English unless you are isolating a specific quantified formula. $\endgroup$
    – Caleb Stanford
    Nov 4, 2016 at 15:12
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    $\begingroup$ Possible duplicate of Sum of radicals greater than 1 $\endgroup$
    – Martin R
    Oct 14, 2019 at 11:01

1 Answer 1

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Hint: By Bernoulli's inequality, $$\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \geqslant \frac1{1+m/n}+ \frac1{1+n/m} = 1$$

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    $\begingroup$ +1 ... Bernoulli's Inequality is one of those undervalued treasures. $\endgroup$
    – Mark Viola
    Nov 4, 2016 at 15:26
  • $\begingroup$ @macavity : Is this true for all integers ? $$\sqrt2=\sqrt{1+1}\\n=2,m=1 \to \\\sqrt{1+1}\geq 1+\frac12=1.5\\\sqrt2=1.41\geq 1.5$$ How it's possible ? $\endgroup$
    – Khosrotash
    Mar 23, 2017 at 18:32
  • $\begingroup$ @Khosrotash you have the inequality the wrong way round. For any natural $n,m$ we have $\sqrt[n]{1+m}\leqslant 1+m/n$. So when you take reciprocals, the inequality reverses. $\endgroup$
    – Macavity
    Mar 23, 2017 at 18:43

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