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The question:

Prove that $x^2+y^2+z^2 ≥ xy+yz+xz$ for all real numbers $x$, $y$ and $z$.

This problem has been posed before, but my question is whether my proof below is correct, since it seems the other answers to this problem are different.

If $x$, $y$ and $z$ are real numbers then $(x-y-z)^2 \geqslant 0$. That is $x^2 + y^2 +z^2 -2xy-2xz-2yz \geqslant 0$. But this implies that $\frac{x^2 + y^2 +z^2}{2} \geqslant xy + xz + yz$, so because $x^2 + y^2 +z^2 \geqslant \frac{x^2 + y^2 +z^2}{2} \geqslant xy + xz + yz$ we get the desired result.

wrb98
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3 Answers3

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Your concerns were already addressed in the comments, so there is no need to keep underlying your actual sign mistake. But on a meta-mathematical point of view, something should have struck you, since your method would have derived a symmetric inequality from a non-symmetric one, kind of strange.

A working and fast proof: $$\color{red}{0\leq} (x-y)^2+(x-z)^2+(y-z)^2 = 2\color{red}{\left(x^2+y^2+z^2-xy-xz-yz\right)}. $$

Jack D'Aurizio
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  • (-1) The OP did not ask for a proof. – TMM Oct 01 '16 at 00:14
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    @TMM: but since OP's proof is not a proof as already remarked in the comments (there is a sign mistake), I guess this answer does not harm. – Jack D'Aurizio Oct 01 '16 at 00:16
  • If you update your answer so you actually address the OP's question (instead of ignoring his question and just giving a different proof), I will remove the downvote. But a "fast answer" which does not even bother to read the whole question is a (-1) for me. – TMM Oct 01 '16 at 00:23
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    That is a downvote based on a wrong impression. OP's concerns were already addressed, there is no need to keep underlying that oversight. – Jack D'Aurizio Oct 01 '16 at 00:26
  • "This problem has been posed before, but my question is whether my proof below is correct, since it seems the other answers to this problem are different." - It is pretty clear to me that (1) the OP's question is purely to see if his proof is correct, and (2) the OP knows that other answers exist (most likely identical to yours), but is not asking for those different answers. In any case, I will keep the downvote on your answer in its current form. – TMM Oct 01 '16 at 00:36
  • But still, answers are not for the OP only. Otherwise, the word community would have no actual meaning here. My opinion is that useful answers should not be downvoted, ever. But you are clearly free to disagree, and I am thankful to you since you explained your downvote, rare thing these days. – Jack D'Aurizio Oct 01 '16 at 00:47
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another way $$xy+yz+zx\le \sqrt { \left( x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) \left( x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 } \right) } =x^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }\\ \\ \\ $$

haqnatural
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Based on hagnatural's comment: No, your proof is incorrect since one of your signs is wrong:

$$(x-y-z)^2 = x^2 + y^2 + z^2 - 2xy - 2xz \color{red}{+} 2yz.$$

And unfortunately there is no direct term $(\pm x \pm y \pm z)^2$ which gives you opposite signs on the cross terms, compared to the squares. (But as noted in the other answers, it is possible to use similar ideas by combining the three terms $(x - y)^2$, $(x - z)^2$, and $(y - z)^2$.)

Community
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TMM
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