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

What is the smallest possible value of $f$

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

2 Answers 2

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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 :)

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

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  • $\begingroup$ Thanks! Brilliant solution. $\endgroup$
    – ggkkll
    Nov 25, 2017 at 20:52
  • $\begingroup$ @ggkkll You are welcome! $\endgroup$ Nov 25, 2017 at 21:58

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