2
$\begingroup$

This question already has an answer here:

$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.

$\endgroup$

marked as duplicate by Martin R, Stefan4024, Vlad, Macavity inequality Apr 27 '16 at 12:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

5
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ +1 Yes. Simple and elegant, although well-known :) $\endgroup$ – almagest Apr 27 '16 at 11:36
0
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.