Problem 41

Algebra Level 4

$\large S = \sqrt{x^2+\dfrac{1}{y^2}}+\sqrt{y^2+\dfrac{1}{z^2}}+\sqrt{z^2+\dfrac{1}{x^2}}$

Given that $x,y$ and $z$ are non-zero reals. Let the minimum value of $S$ be $S_{\min}$ . Find $(S_{\min})^2$ .

This problem has been taken from Azerbaijan TST prepare for IMO 2016.

For more inequality problems, see this set: It's all about the inequality .

The answer is 18.

1 solution

P C
Apr 29, 2016

Call the expression P, we can apply Minkowski Inequality directly $P\geq\sqrt{(x+y+z)^2+\bigg(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\bigg)^2}$ By Titu's Lemma $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq\frac{9}{x+y+z}$ $\therefore P\geq \sqrt{(x+y+z)^2+\frac{9^2}{(x+y+z)^2}}\geq\sqrt{18} \ (By \ AM-GM)$ So $S_{min}^2=18$ , the equality holds when $a=b=c=\pm{1}$

Nice... Did other way reaches at same answer.

Pawan pal - 5 years, 1 month ago

I see a good application of Jensen's Inequality followed by AM-GM.

Sal Gard - 5 years, 1 month ago

How does your answer follow from

https://brilliant.org/discussions/thread/proof-of-minkowskis-inequality/

D S - 3 years, 6 months ago

like woulnt int be sqrt(x+y+z+1/x+1/y+1/z)