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Prove that for every $n,m \in \Bbb N $ $$ \frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1 $$

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2 Answers 2

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According to Bernoulli's inequality $$(1+m)^\frac{1}{n} \leq 1+\frac{m}{n}, \\ (1+n)^\frac{1}{m} \leq 1+\frac{n}{m},$$ so that $$\frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \geq \frac{1}{1+\frac{m}{n}}+\frac{1}{1+\frac{n}{m}}=1. $$

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For every $a,b \in \mathbb{N}$ we have $$ \left(1+\frac{b}{a}\right)^a=\sum_{i=0}^a{a\choose i}\left(\frac{b}{a}\right)^i=1+b+\ldots \ge 1+b. $$ Therefore $$ (1+b)^{1/a}\le 1+\frac{b}{a}=\frac{a+b}{a} \quad \forall a,b \in \mathbb{N}. $$ Hence for every $m,n \in \mathbb{N}$ we have $$ \frac{1}{\sqrt[n]{1+m}}+\frac{1}{\sqrt[m]{1+n}} \ge \frac{n}{m+n}+\frac{m}{m+n}=1. $$

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