Home > IMO, Inequalities, Olympiad > Balkan Mathematical Olympiad 2011 Problem 2

Balkan Mathematical Olympiad 2011 Problem 2


Given real numbers x,y,z such that x+y+z=0, show that
\dfrac{x(x+2)}{2x^2+1}+\dfrac{y(y+2)}{2y^2+1}+\dfrac{z(z+2)}{2z^2+1}\ge 0
When does equality hold?

BMO 2011 Problem 2

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Categories: IMO, Inequalities, Olympiad Tags: , ,
  1. May 18, 2011 at 7:10 am

    \frac{x(x+2)}{2x^2+1} +\frac{1}{2}=\frac{(2x+1)^2}{2(2x^2+1)} .

    \sum_{cyc}\frac{(2x+1)^2}{2x^2+1}\ge 3 .

    Assuming xy \ge 0 we have

    \sum_{cyc}\frac{(2x+1)^2}{2x^2+1} \ge  \frac{(2x+2y+2)^2}{2(x^2+y^2)+2}+\frac{(2z+1)^2}{2z^2+1}\ge \frac{2(1-z)^2}{z^2+1}+\frac{(2z+1)^2}{2z^2+1} =3 + \frac{2z^2(z-1)^2}{(z^2+1)(2z^2+1)}\ge  3.

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