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Show that:

$$\|{x-y}\|\|x+y\| \leq \|x\|^2+\|y\|^2$$

My guess is that we need to show that the left-hand side reduces to $\|x+y\|^2$ (the Triangle Inequality holds) but I haven't been able to make much progress in doing so.

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Well, the norm is multiplicative, so we can use triangle inequality directly. – awllower Feb 8 '13 at 3:45
@awllower The norm is multiplicative? – 1015 Feb 8 '13 at 22:32
Oh, I mistook this norm as the norm over an algebraic number field, where the norm is certainly multiplicative. The situation here, however, is quite different, and indeed to prove this inequality it is essential to avail of parallelogram law. Thanks for the attention. – awllower Feb 9 '13 at 13:55
up vote 4 down vote accepted

Hint: Use $ab\leq (a^2+b^2)/2$ and the parallelogram law (

Ok, I'll expand now.

Since $(a-b)^2\geq 0$, we have $a^2-2ab+b^2\geq 0$ hence $ab\leq (a^2+b^2)/2$ for all $a.b\in\mathbb{R}$.

So $$ \|x-y\|\|x+y\|\leq\frac{1}{2}(\|x-y\|^2+\|x+y\|^2). $$

Now compute $$ \|x-y\|^2+\|x+y\|^2=(x-y,x-y)+(x+y,x+y)=2\|x\|^2+2\|y\|^2. $$ This is called the parallelogram law.

I think you can conclude.

Note this result is true in a general inner product space.

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Hint: The inequality is equivalent to $$\|x-y\|^2\|x+y\|^2\leq (\|x\|^2+\|y\|^2)^2.$$

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Can one of you tell me how this leads to the desired inequality? I really can't see it. And I can't see how you would do without using at some point the parallelogram law. – 1015 Feb 8 '13 at 4:08
Since it is involved only with positive numbers, squaring leads to an equivalent inequality. Now $\|x-y\|²=<x-y,x-y>=\|x\|²+\|y\|²-2<x,y>$ and $\|x+y\|²=<x+y,x+y>=\|x\|²+\|y\|²+2<x,y>$, so the inequality follows. – awllower Feb 9 '13 at 14:01
@awllower Oh, I see. So we can actually go around the parallelogram law. Interesting. Thanks. – 1015 Feb 11 '13 at 1:04
You are welcomed. – awllower Feb 11 '13 at 3:00

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