Is there an inequality that $\|a+b\|^2 \le \|a\|^2 + \|b\|^2$?

Question

For a norm $\|\cdot\|$ by triangle inequality we have $\|a+b\| \le \|a\| + \|b\|$ and by Cauchy–Schwarz inequality $\|ab\| \le \|a\|\|b\|$, but I am not sure if the following inequality holds always true: $\|a+b\|^2 \le \|a\|^2 + \|b\|^2$.

Thanks for the answers!

Then, let's say $\|a+b\|^2 \lesssim \|a\|^2 + \|b\|^2$, is that true?

Answer 1

If we have an inner product $\langle . | . \rangle$, then: $$ ||a+b||^2 = \langle a+b | a+b \rangle = \langle a | a \rangle +2\langle a | b \rangle + \langle b | b \rangle = ||a||^2+||b||^2+2\langle a | b \rangle $$ Thus your assertion does not hold for say: $a=\alpha b$ with $\alpha>0$.

Answer 2

No, take $a=b=1$. Then you get $\|a+b\|^2=4$ and $\|a\|^2=\|b\|^2=1$, so $4\le 1+1$ does not hold.

Answer 3

If $a=b\neq 0$ then $||a+b\|^2=4\|a\|^2\neq \|a\|^2+\|a\|^2=2\|a\|^2$.

But you could prove that $x\to \|x\|^2$ is a convex function, i.e. $\|sa+tb\|^2\leq s\|a\|^2+t\|b\|^2$ whenever $s+t=1$ and $ s,t\geq 0$.

So in particular if $s=t=\frac 12$ then $\|a+b\|^2\leq 2(\|a\|^2+\|b\|^2)$