Triangle inequality squared?

Question

I am in the process of understanding a proof. There, for example, the following is indicated:

$$\big\||a\rangle+|b\rangle\big\|^2\leq\big\||a\rangle\big\|^2+2\big\||a\rangle\big\|\big\||b\rangle\big\|+\big\||b\rangle\big\|^2$$

This should now be used in the proof (as a tool for further calculation). This reminds me of the triangle inequality, is that the same? or in what do the expressions differ?

Then I would be interested, assuming it would be: $$\big\||a\rangle-|b\rangle\big\|^2$$ What would be the result of this?

Maybe mentioned by the way: The Ket notation was used because the calculation is one from the topic of quantum computing.

PS: I am not sure if the title of the question is well chosen, I did not know how to put it another way. So if someone has a better title then I would improve the title.

Answer 1

The triangle inequality gives $\Vert|a\rangle\pm|b\rangle\Vert\le\Vert a\Vert+\Vert b\Vert$ for both choices of the $\pm$ sign. Squaring gives $\Vert|a\rangle\pm|b\rangle\Vert^2\le\Vert a\Vert^2+2\Vert|a\rangle\Vert\Vert|b\rangle\Vert+\Vert b\Vert^2$.

Answer 2

Disclaimer: the ket notation is painful to watch.

I think it's the ket notation that it not allowing you to see the forest for the trees. From the triangle inequality: $$ \|x+y\|\leq\|x\|+\|y\|. $$ Now square, and expand the binomial on the right: $$ \|x+y\|^2\leq(\|x\|+\|y\|)^2=\|x\|^2+2\|x\|\,\|y\|+\|y\|^2. $$ Actually, this is the way one usually proves the triangle inequality via Cauchy-Schwarz: $$ \|x+y\|^2=\|x\|^2+2\operatorname{Re}\langle x,y\rangle+\|y\|^2\leq\|x\|^2+2\|x\|\,\|y\|+\|y\|^2=(\|x\|+\|y\|)^2. $$

The minus sign on the left makes no difference, as the triangle inequality gives $$ \|x-y\|=\|x+(-y)\|\leq\|x\|+\|-y\|=\|x\|+\|y\|. $$