Prove the following inequality:

Prove the following inequality:?
|x + y + z|^2 is less than or equal to 3 * (x^2 + y^2 + z^2).

The | |'s are absolute values.

Note that
|x + y + z|^2 ≤ 3(x^2 + y^2 + z^2)
<==> (x^2 + y^2 + z^2) + 2(|xy| + |xz| + |yz|) ≤ 3(x^2 + y^2 + z^2)
<==> |xy| + |xz| + |yz| ≤ x^2 + y^2 + z^2.

To obtain this last inequality, use the Arithmetic Mean-Geometric Mean Inequality:
(x^2 + y^2)/2 ≥ √(x^2 * y^2) = |xy|.

Similarly,
(y^2 + z^2)/2 ≥ |yz|.
(x^2 + z^2)/2 ≥ |xz|.

Adding these three inequalities yields |xy| + |xz| + |yz| ≤ x^2 + y^2 + z^2, as desired.

I hope this helps!