<s>[INST] Maximal Smallest Number Ran into a interesting problem which I have no idea how to solve but have the desire to. 

Let a and b be two positive real numbers and let $m$(a,b) be the smallest of the three numbers $a,$ $1/b$ and $1/a + b.$ For which pairs (a,b) is $m$(a,b) maximal? Prove!

 [/INST] It is not too hard to see that the maximum can only occur for $a=\frac{1}{b}=\frac{1}{a}+b$.
Suppose for instance that $a<\min(\frac{1}{b},b+\frac{1}{a})$, then we can increase $a$ a little and improve our $m(a,b)$.
For $\frac{1}{b}$ and $b+\frac{1}{a}$ similar things happen.
Thus $a=\frac{1}{b} = b+\frac{1}{a}$ implies $a = \sqrt{2}$ and $b = \frac{1}{2}\sqrt{2}$.
Checking the answer is quite easy:
Suppose that $m(a,b)>\sqrt{2}$, then $a>\sqrt{2}$ and $\frac{1}{b}>\sqrt{2}$. But then $\frac{1}{a}<\frac{1}{2}\sqrt{2}$ and $b<\frac{1}{2}\sqrt{2}$, so $b+\frac{1}{a}<\sqrt{2}$. A contradiction.
Thus $m(a,b)\leq\sqrt{2}$ for all $a$ and $b$. We already know that $m(\sqrt{2},\frac{1}{2}\sqrt{2})=\sqrt{2}$, thus this is optimal.
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