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

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

marked as duplicate by Martin R, Stefan4024, Vlad, Macavity inequality StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 27 '16 at 12:14

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.

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

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

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