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 5 of 5: Chain two classical facts, using xyz≥1xyz\ge1
Previous stepNext step
In plain words

The first inequality is the standard fact ∑x2≥∑xy\sum x^2\ge\sum xy; the second turns xy+yz+zxxy+yz+zx into xyzxyz times ∑1x\sum\frac1x, and xyz≥1xyz\ge1 is exactly the hypothesis needed to keep that quantity at least ∑1x\sum\frac1x.

x2+y2+z2 ≥ xy+yz+zx=xyz(1x+1y+1z) ≥ 1x+1y+1zx^2+y^2+z^2\ \ge\ xy+yz+zx = xyz\left(\frac1x+\frac1y+\frac1z\right)\ \ge\ \frac1x+\frac1y+\frac1z
Detailed analysis

The inequality x2+y2+z2≥xy+yz+zxx^2+y^2+z^2\ge xy+yz+zx holds for all reals (it is equivalent to 12[(x−y)2+(y−z)2+(z−x)2]≥0\frac12\left[(x-y)^2+(y-z)^2+(z-x)^2\right]\ge0). Also xy+yz+zx=xyz(1z+1x+1y)xy+yz+zx=xyz\left(\frac1z+\frac1x+\frac1y\right) identically, and since xyz≥1xyz\ge1 and 1x+1y+1z>0\frac1x+\frac1y+\frac1z>0, we get xyz(1x+1y+1z)≥1x+1y+1zxyz\left(\frac1x+\frac1y+\frac1z\right)\ge \frac1x+\frac1y+\frac1z. Chaining the two gives the inequality of the previous step, which by Step 3 completes the proof; equality holds at x=y=z=1x=y=z=1.