x,y,z >= 0 .. Prove :
If at least one of x, y, z is zero, the statement is obviously true:
Let's say x=0, we get:
which, after factoring gives:
So we consider the case where all three x,y,z are greater than zero. Assume further that x < y < z and that y = l x and z = kx for some l, k > 0. Note that k > l.
Now we write the inequality in terms of x, l, and k:
and move all terms to the left:
Divide both sides by x^2, since x^2 > 0 this will not change the direction of the inequality we are trying to prove:
Note that (Cauchy inequality?)
The above result comes from 'arithmetic mean' is greater than or equal to the geometric mean, formally:
So it suffices to show that:
Moving all terms to the left hand side:
Because l < k , all the terms on the left hand side are positive.
I hope this helps.