# Trouble proving natural number inequality $\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \ge 1$

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?

• For all natural numbers $m$, $n$? – 6005 Nov 4 '16 at 15:07
• yes. I forgot to add that – Joe Shmoe Nov 4 '16 at 15:08
• @6005 isn't there a $\forall$? – Joe Shmoe Nov 4 '16 at 15:10
• 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. – 6005 Nov 4 '16 at 15:12
• maybe use induction? – Mastrem Nov 4 '16 at 15:16

## 1 Answer

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

• +1 ... Bernoulli's Inequality is one of those undervalued treasures. – Mark Viola Nov 4 '16 at 15:26
• @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 ? – Khosrotash Mar 23 '17 at 18:32
• @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. – Macavity Mar 23 '17 at 18:43