# 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.

• For all natural numbers $m$, $n$? Nov 4, 2016 at 15:07
• yes. I forgot to add that Nov 4, 2016 at 15:08
• @6005 isn't there a $\forall$? Nov 4, 2016 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. Nov 4, 2016 at 15:12
• Possible duplicate of Sum of radicals greater than 1 Oct 14, 2019 at 11:01

Hint: By Bernoulli's inequality, $$\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \geqslant \frac1{1+m/n}+ \frac1{1+n/m} = 1$$
• @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 ? Mar 23, 2017 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. Mar 23, 2017 at 18:43