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

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

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## 3 Answers

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