Proof help: Prove that $x^2+y^2+z^2 \geq xy+xz+yz$

Question

$x^2+y^2+z^2 \geq xy+xz+yz $ for all real numbers, x, y, and z.

I'm not very good with working inequality proofs. Can someone help me prove this? The technique doesn't really matter.

score 5 · Answer 1 · answered Apr 27 '16 at 11:34

5

First observe that $ (x-y)^2 + (x-z)^2 + (y-z)^2 \geq 0 $.

Next, expand the LHS to obtain:

$ 2x^2 + 2y^2 + 2z^2 - 2xy - 2xz - 2yz \geq 0 $

Now you simply divide by two and add $xy + xz + yz$ to both sides.

answered Apr 27 '16 at 11:34

J. Young

score 1 · Answer 2 · answered Apr 27 '16 at 11:33

1

Hint:$ (x-y)^2>0$.Can you take it from here?

answered Apr 27 '16 at 11:33

DeepSea

+1 Yes. Simple and elegant, although well-known :) – almagest Apr 27 '16 at 11:36

score 0 · Answer 3 · answered Apr 27 '16 at 11:40

0

Hint complete square to get $(x+y+z)^2\geq 3(xy+zy+xz)+2xyz$ and then use AM-Gm on rhs

answered Apr 27 '16 at 11:40

Archis Welankar

3 Answers3