Question

prove that x^3+y^3+z^3_3xyz

Answer 1

x,y,z=>x3,y3,z3>0

Now As you know A.M≥G.M

AM of x3,y3,z3=x3+y3+z33  

G.M of x3,y3,z3=(x3∗y3∗z3)1/3

A.M≥G.M

x3+y3+z33 ≥(x3∗y3∗z3)1/3−−−(1)

solving (1) makes it

x3+y3+z3≥3xyz