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 1 of 5: Compare each term to a simpler fraction
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In plain words

The difference between the original term and a version with a friendlier denominator turns out to be a perfect square times a nonnegative quantity, so it never has the wrong sign.

x5−x2x5+y2+z2−x5−x2x3(x2+y2+z2)=x2(x3−1)2(y2+z2)(x5+y2+z2) x3(x2+y2+z2)≥0\frac{x^5-x^2}{x^5+y^2+z^2}-\frac{x^5-x^2}{x^3(x^2+y^2+z^2)}=\frac{x^2(x^3-1)^2(y^2+z^2)}{(x^5+y^2+z^2)\,x^3(x^2+y^2+z^2)}\ge0
Detailed analysis

Direct algebra shows the difference of the two fractions equals x2(x3−1)2(y2+z2)(x5+y2+z2) x3(x2+y2+z2)\dfrac{x^2(x^3-1)^2(y^2+z^2)}{(x^5+y^2+z^2)\,x^3(x^2+y^2+z^2)}, whose numerator and denominator are both nonnegative for x,y,z>0x,y,z>0. Hence x5−x2x5+y2+z2≥x5−x2x3(x2+y2+z2)\dfrac{x^5-x^2}{x^5+y^2+z^2}\ge\dfrac{x^5-x^2}{x^3(x^2+y^2+z^2)} for every x,y,z>0x,y,z>0, with no need for xyz≥1xyz\ge1 yet.