My two cents on this problem:

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.