Proving a certain inequality in $\mathbb{R}^n$

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? –  julien 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

Hint: Use $ab\leq (a^2+b^2)/2$ and the parallelogram law (http://en.wikipedia.org/wiki/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 $$\|x-y\|^2\|x+y\|^2\leq (\|x\|^2+\|y\|^2)^2.$$
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