IMO 2005 · Problem 3

Let \(x,y,z\) be three positive reals such that \(xyz\geq 1\). Prove that \[ \frac{x^5-x^2}{x^5+y^2+z^2} +\frac{y^5-y^2}{x^2+y^5+z^2} +\frac{z^5-z^2}{x^2+y^2+z^5}\geq 0. \]

Official text of the problem. Official problem paper (PDF)

Solution

Put \(S=x^2+y^2+z^2\). The key observation is the identity \[ \frac{x^5-x^2}{x^5+y^2+z^2} -\frac{x^2-\frac1x}{S} = \frac{(y^2+z^2)(x^3-1)^2} {xS(x^5+y^2+z^2)} \geq 0. \] All denominators are positive. Adding this inequality and its two cyclic counterparts gives \[ \frac{x^5-x^2}{x^5+y^2+z^2} +\frac{y^5-y^2}{x^2+y^5+z^2} +\frac{z^5-z^2}{x^2+y^2+z^5} \geq \frac{S-\left(\frac1x+\frac1y+\frac1z\right)}{S}. \]

Now \(xyz\geq1\) implies \[ \frac1x+\frac1y+\frac1z =\frac{xy+yz+zx}{xyz} \leq xy+yz+zx \leq x^2+y^2+z^2=S, \] where the last inequality follows from \[ x^2+y^2+z^2-xy-yz-zx =\frac12\bigl((x-y)^2+(y-z)^2+(z-x)^2\bigr)\geq0. \] Thus the required sum is nonnegative.

Equality holds precisely when \(x=y=z=1\): equality in the three initial inequalities forces \(x^3=y^3=z^3=1\), and this triple satisfies the original inequality with equality.

The solution was checked against Evan Chen's published IMO solution notes.