What is the geometric interpretation of $ \frac{1}{2}||a||^2\leq \langle a,b\rangle $?

Question

Give $a, b$ in $\mathbb{R}^n$. What is the geometric interpretation of the following?

$$ \frac{1}{2}\|a\|^2 \leq \langle a,b\rangle $$

In other words, what criteria should $a$ and $b$ have to satisfy the above inequality?

If possible, please provide 2D interpretation when $n=2$.

Answer 1

If $a$ is $0$, then any $b$ will satisfy the inequality.

If $a$ is different from $0$, then it is a bit more interesting. We can draw a line segment from $0$ to $a$ and perpendicularly bisect it like so:

This line divides the plane into two regions. All the vectors $b$ which are on the same side as $a$ (the shaded region) will satisfy the inequality, and the others won’t.

I will leave it up to you to formalize this argument for $\mathbf{R}^n$ in terms of orthogonal subspaces, but this picture can serve as a motivation.

Answer 2

There's an equivalent interpretation in terms of distances.

The condition $\displaystyle \frac{1}{2}\|a\|^2 \leq \langle a,b\rangle$ is equivalent to $\|a\|^2 - 2\langle a,b\rangle \leq 0$. Observe that $$ \begin{aligned} \|a-b\|^2 &= \langle a - b, a - b\rangle \\ &= \langle a,a \rangle - 2\langle a,b\rangle + \langle b, b\rangle \\ &= \|a\|^2 - 2\langle a,b \rangle + \|b\|^2 \end{aligned}$$ Therefore $\|a\|^2 - 2 \langle a,b \rangle \leq 0$ if and only if $\|a-b\|^2 \leq \|b\|^2$, if and only if $\|a-b\| \leq \|b\|$.

So, $a$ and $b$ satisfy the condition if and only if $b$ is closer to $a$ than it is to the origin. This is an equivalent characterization to the one given in the answer by @shoteyes.