# Inequality question---Max and Min

$f(x,y)=\max\left\{x,y,\frac 1x + \frac 1y\right\}$, where x and y are positive real numbers.

What is the smallest possible value of $f$

• Have you made any progress on this question?
– user499203
Commented Nov 25, 2017 at 17:04
• I've tried x=y, so the smallest value would be root 2. Now, I'm considering x>y. Commented Nov 25, 2017 at 17:07

Assume without loss of generality that $x\geq y$. Then the question is equivalent to finding the minimum value of $max (x,\frac {1}{x}+\frac {1}{y}))$. Now, because of the new assumption, we get $1/x+1/y\geq 2/x$so the minimum value would be of the form $max (x,2/x)$. I leave it to you to calculate the exact minimum value :)

• i think you must consider all possible cases Commented Nov 25, 2017 at 17:12
• and what is if $$x<y$$? Commented Nov 25, 2017 at 17:24
• @Dr.SonnhardGraubner That's obvious: $f(x,y)=f(y,x)$ and $y>x$ so already covered.
– hvd
Commented Nov 25, 2017 at 17:26
• Thanks! @Or Kedar Commented Nov 25, 2017 at 20:53

Let $f(x,y)=k$.

Thus, $x\leq k$, $y\leq k$ and $$\frac{2}{k}\leq\frac{1}{x}+\frac{1}{y}\leq k,$$ which gives $$k^2\geq2$$ or $$k\geq\sqrt2.$$ The equality occurs for $$x=y=\frac{1}{x}+\frac{1}{y}=\sqrt2,$$ which says that $\sqrt2$ is a minimal value.

• Thanks! Brilliant solution. Commented Nov 25, 2017 at 20:52
• @ggkkll You are welcome! Commented Nov 25, 2017 at 21:58