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 2 of 5: Rewrite the simpler bound
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x5−x2x3(x2+y2+z2)=x2−1xx2+y2+z2\frac{x^5-x^2}{x^3(x^2+y^2+z^2)}=\frac{x^2-\frac1x}{x^2+y^2+z^2}
Detailed analysis

Dividing the numerator and denominator of the right-hand side of the previous step by x3x^3 gives x5−x2x3(x2+y2+z2)=x2−1xx2+y2+z2\dfrac{x^5-x^2}{x^3(x^2+y^2+z^2)}=\dfrac{x^2-\frac1x}{x^2+y^2+z^2}, since x5/x3=x2x^5/x^3=x^2 and x2/x3=1/xx^2/x^3=1/x.