MathLabs

Problem 3

Let x,y,z>0x, y, z > 0 satisfy xyz≥1xyz\ge 1. Prove that x5−x2x5+y2+z2+y5−y2x2+y5+z2+z5−z2x2+y2+z5≥0.\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} \ge 0.
Step 3 of 5: Add the three bounds
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In plain words

All three lower bounds share the denominator x2+y2+z2x^2+y^2+z^2, so adding them cyclically combines neatly into one fraction.

∑cycx5−x2x5+y2+z2 ≥ (x2+y2+z2)−(1x+1y+1z)x2+y2+z2\sum_{cyc}\frac{x^5-x^2}{x^5+y^2+z^2}\ \ge\ \frac{(x^2+y^2+z^2)-\left(\frac1x+\frac1y+\frac1z\right)}{x^2+y^2+z^2}
Detailed analysis

Summing the bound from Steps 1–2 over the cyclic shift x→y→z→xx\to y\to z\to x, and noting the three lower bounds all share the denominator x2+y2+z2x^2+y^2+z^2, gives this combined lower bound for the left-hand side of the problem.