Q:

Find max f(a,b)

$f(a,b)=min(a,b,\frac1a+\frac1b)$, where $a$ and $b$ are positive real numbers.
Basically what I would do is simply find when are these 3 terms equal($a=b=\sqrt 2$), and then manually check what happens when I up/down both values and that should show that $max(f)=\sqrt 2$, but I also think there should be a more elegant way to do this, so all ideas are welcome.

A:

Because $f(a,b)$ is symmetric in $a,b$, it can be assumed WLOG that $a \le b\,$, in which case $f(a,b)=\min(a, \frac{1}{a}+\frac{1}{b})\,$. Since $\frac{1}{a}+\frac{1}{b} \le \frac{2}{a}\,$ it follows that $f(a,b) \le \min(a, \frac{2}{a})\,$, with equality iff $b=a\,$. Out of the two numbers $a,\frac{2}{a}$ which multiply together to $2\,$, at least one must be smaller than or equal to $\sqrt{2}\,$, so $f(a,b) \le \sqrt{2}\,$ which is to say that $\sqrt{2}$ is an upper bound for $f(a,b)\,$. Since that upper bound is in fact attained for $a=b=\sqrt{2}\,$ it follows that $\sqrt{2}$ is an actual maximum.

