How do I prove that $\min A = \frac{1}{3}$ for: $\left\{m,n\in\mathbb{N}:\frac{mn}{1+m+n}\right\}$
Prove that $\frac{mn}{1+m+n}\ge\frac{1}{3}$ for $m,n\in\mathbb{N}$
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real-analysis
2 Answers
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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 $1/3$.