Proving $\|a + b\|^2 \geq \frac{1}{2}\|a\|^2 - \|b\|^2$ for $a$, $b$ in $\Bbb{R}^d$

Question

I have encountered the following inequality in the optimization article.

For all $a, b \in \mathbb{R}^d$ the following inequality holds: $\|a + b\|^2 \geq \frac{1}{2}\|a\|^2 - \|b\|^2$.

The norm is a simple $l_2$-norm. This is all the information that was given.

It was written in the Appendix and was marked as "basic" inequality, without any sketch of proof or even naming. I have tried to prove it on my own or to find anything related to it, and failed.

Could you please explain to me how to prove it?

Answer 1

See the parallelogram law. \begin{align} \|a\|^2 + \|a+2b\|^2 &= \|(a+b)-b\|^2 + \|(a+b)+b\|^2 \\ &= 2\|a+b\|^2 + 2\|b\|^2. \end{align} Dividing by $2$ and noting $\|a+2b\|^2\ge 0$ yields your equality.