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

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: $\|a+b\|^2 \le \|a\|^2 + \|b\|^2$.

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

• Take $a=b$ non zero. Apr 19 '18 at 13:57
• Since $\|a+b\|^2=\|a\|^2+2a\cdot b\|b\|^2$, that inequality hinges on if $a\cdot b$ is positive or negative.
– anon
Apr 19 '18 at 13:58
• $\| a+b \|^2\leq 2(\|a\|^2+\|b\|^2)$ Apr 19 '18 at 14:19
• Thanks! I think I should use "$\lesssim$" instead of "$\le$" as in the textbook did Apr 19 '18 at 14:31

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 $\alpha>0$.

• So $\|a+b\|^2 \le 2(\|a\|^2 + \|b\|^2)$ is true? Apr 19 '18 at 14:25
• Yes because then you would have $2 \langle a|b \rangle \leq \langle a|a \rangle + \langle b|b \rangle$ i.e. $0\leq \langle b-a|b-a \rangle = ||b-a||^2$ which holds Apr 19 '18 at 14:38
• Thank you very much! By the way, can we say $\|ab\|\le\|a\|\|b\|$ holds by Cauchy–Schwarz inequality? I saw Cauchy–Schwarz inequality in textbook is always an inner product form $|<a,b>|$ on LHS. Apr 19 '18 at 16:10
• No it is not possible, that would be a normed algebra. $|\langle a,b \rangle|$ and $||ab||$ are nothing alike and actually, in a vector space, $ab$ makes no sense at all since there is no multiplication in vector spaces. Apr 19 '18 at 16:14
• Well. Assume $a$, $b$ are two functions integrable over the domain $\Omega$ for the norm, then can we say $\|a b\|_{\Omega} \le \|a\|_{\Omega} \|b\|_{\Omega}$, where $ab$ means the product of the two functions? Apr 20 '18 at 11:48

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.

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 $\|a+b\|^2\leq 2(\|a\|^2+\|b\|^2)$