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How do I prove that $\min A = \frac{1}{3}$ for: $$\left\{m,n\in\mathbb{N}:\frac{mn}{1+m+n}\right\}$$

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2 Answers 2

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It is same as proving $1+m+n \leq 3mn$

Same as proving $\frac{1}{m}+\frac{1}{n}+\frac{1}{mn} \leq 3$

Since $m,n \geq 1$,

$\Rightarrow \frac{1}{m} \leq 1, \frac{1}{n} \leq 1, \frac{1}{mn} \leq 1$

Thus the result follows. Equality holds when $m=n=1$.

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Another approach is the following:

$\frac{mn}{1+m+n}=\frac{2+mn}{1+m+n}-\frac{2}{1+m+n}\geq1-\frac{2}{1+m+n}\geq 1-\frac{2}{3}=\frac{1}{3} $

Note that the first inequality follows since $(1-m)(1-n)\geq 0$ for $n$ and $m$ greater (or equal) than $1$.

Equality is obtained when $m=n=1$, thus the minimum is $1/3$.

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