86. A very nice maxmin.

by Virgil Nicula, Aug 16, 2010, 1:43 AM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=358595&p=1984871#p1984871
Quote:
Denote $m(x,y)=\min \left\{x,y,\frac 1x+\frac 1y\right\}$ for any $x>0$ and $y>0$ . Find $M=\max_{\{x,y\}\subset (0,\infty)}m(x,y)$ .
Proof. Observe that for any $x>0$ and $y>0$ have $\left\|\begin{array}{ccc} m(x,y)\le x & \implies & \frac 1x\le \frac {1}{m(x,y)}\\\ m(x,y)\le y & \implies & \frac 1y\le \frac {1}{m(x,y)}\end{array}\right\|\implies$

$m(x,y)\le$ $\frac 1x+\frac 1y\le \frac {2}{m(x,y)}\implies$ $m(x,y)\le \sqrt 2=m\left(\sqrt 2,\sqrt 2\right)$ . In conclusion, $M=\sqrt 2$ .
Quote:
Generalization (Training Team Vietnam 2009). $x_k>0\ ,\ k\in\overline{1,n}$ and $S_n=\min\left\{x_1,x_2+\frac{1}{x_1},\ldots ,x_n+\frac{1}{x_{n-1}},\frac{1}{x_n}\right\}$ $\implies$ $\max S_n=2\cos{\frac{\pi}{n+2}}$ .
This post has been edited 10 times. Last edited by Virgil Nicula, Nov 23, 2015, 2:14 PM

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