# Math Help - Prove

1. ## Prove

x,y,z >= 0 .. Prove :

2. Originally Posted by dhiab
x,y,z >= 0 .. Prove :
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:
$y^2 + z^2 + 1 \geq 2yz$
which, after factoring gives:

$(y-z)^2 + 1 \geq 0$

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:

$x^2 + (lx)^2 + (kx)^2 + 2 x (lx)(kx) + 1 \geq 2(xlx + xkx + lxkx)$

and move all terms to the left:

$x^2 + l^2 x^2 + k^2 x^2 + 2 klx^3 + 1 \geq 2(x^2 l + x^2 k + lk x^2)$

Divide both sides by x^2, since x^2 > 0 this will not change the direction of the inequality we are trying to prove:

$1 + l^2 + k^2 + 2klx + \frac{1}{x^2} \geq 2l + 2lk + 2k$

Note that $l^2 + k^2 \geq 2lk$ (Cauchy inequality?)

The above result comes from 'arithmetic mean' is greater than or equal to the geometric mean, formally:

$\frac{x+y}{2} \geq \sqrt{xy}$

So it suffices to show that:

$1 + 2klx + \frac{1}{x^2} \geq 2l + 2k$

Moving all terms to the left hand side:

$1+ 2klx + \frac{1}{x^2} - 2 (l-k) \geq 0$

Because l < k , all the terms on the left hand side are positive.

I hope this helps.