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Does the inequality $2 \langle x , y \rangle \leqslant \langle x , x \rangle + \langle y , y \rangle $, where $$ \langle \cdot, \cdot \rangle $$ denotes scalar product, have a name?

I've tried looking at several inequalities on wikipedia but I didn't find this one. And of course googling doesn't work for this purpose.

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This inequality is too obvious, so I cannot imagine that this has a distinguished name. We may identify it as a variant of AM-GM inequality, since both can be written as $|x - y|^2 \geq 0$ (when only two variables are involved). – Sangchul Lee Feb 22 '12 at 17:55
@M. Alaggan: Why did you post this question? What is your motivation? – Beni Bogosel Feb 22 '12 at 21:40
@BeniBogosel: I've a differential privacy mechanism whose error bound I computed through the mean value theorem in the general case, and in an application I've managed to compute the error bound via this "shortcut" instead of having to go through the mean value theorem. This shortcut was found by coincidence, but I am interested to know its name for two reasons: 1) to use its name in my paper and 2) to have more information or insight about why it is true. – M. Alaggan Feb 22 '12 at 21:45
@BeniBogosel: It would be equally good for me to learn the intuition of why this inequality is so obvious, that it doesn't have a name. It was (and still) not immediately obvious to me. – M. Alaggan Feb 22 '12 at 21:50
@M.Alaggan: As far as I think, I don't think that you need a special name for this inequality. Even without a proof it stands out as true, if one thinks a bit of the properties of the scalar properties. Anyway, goodluck with your article :) – Beni Bogosel Feb 22 '12 at 21:52
up vote 6 down vote accepted

It is the development of $$ \langle x-y,x-y \rangle \geq 0$$ and it follows from the positive definitness of the scalar product.

Apart of the above proof of the inequality, and as a response to the comments to the question, here are a few reasons as to why this inequality should be true, at a first glance:

  • a scalar product has the properties of the multiplication on the real line, so the inequality $2xy\leq x^2+y^2$ should pop up while looking at the given inequality;

  • Cauchy Schwarz immediatley implies the inequality: $$2\langle x,y \rangle \leq 2 \|x\|\|y\| \leq\|x\|^2+\|y\|^2 $$

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Yes, of course. But that is not a name. – Harald Hanche-Olsen Feb 22 '12 at 20:34
Usually very simple facts do not have names. It is exactly as if you would ask what is the name for $x^2 \geq 0$. And moreover, this inequality is (one liner) proved to be weaker than least two inequalities which already have names: Cauchy Schwarz and the triangle inequality. – Beni Bogosel Feb 22 '12 at 21:33
I totally agree. So the question, as asked, probably does not have an answer. Which is why I gave an almost-answer instead. I suppose we are both guilty of cheating. – Harald Hanche-Olsen Feb 22 '12 at 21:35
I would not consider anyone who answered this question guilty. I would like to know the motivation of the OP for posting this question. – Beni Bogosel Feb 22 '12 at 21:39

It is essentially Young's inequality.

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Okay, maybe that's stretching it. But it is not far from the truth. – Harald Hanche-Olsen Feb 22 '12 at 17:52

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