How to prove the following norm inequality

Question

If $x$, $y$ and $a$ are vectors in $\mathbb{R}$, is the following inequality true?

$||y - a|| - ||x-a|| \ge (\frac{x-a}{||x-a||})^T(y-x)$

I cannot come up with a counterexample, but I also do not know how to approach proving norm inequalities.

Answer 1

As I have written in my comment, it is enough to prove

\begin{equation}\|y\|-\|x\|\ge \frac{\langle x,y-x\rangle}{\|x\|}\tag{1}\end{equation}

for $x\ne 0.$

By the Schwarz inequality we have $|\langle x,y\rangle|\le\|x\|\cdot\|y\|$, in particular $$\langle x,y\rangle\le\|x\|\cdot\|y\|.$$

Then

$$\langle x,y\rangle-\|x\|^2\le\|x\|\cdot\|y\|-\|x\|^2.$$

Hence

$$\langle x,y\rangle-\langle x,x\rangle \le\|x\|\cdot\|y\|-\|x\|^2.$$

By the properties of the inner product

$$\langle x,y-x\rangle\le\|x\|\cdot\|y\|-\|x\|^2.$$

Dividing both sides by $\|x\|$ we arrive at the desired inequality.

Now, assuming $x\ne a$, apply the inequality $(1)$ to $x-a,y-a$ to get the inequality you have written.