Vector inequality, how to prove it?

Question

Foe a vector $v$ and $a,b$ such that

$ 0 < 1-a-b \leq \| v\|^2 \leq 1-a+b $

with $ 0<a,b<1 $ it holds the inequality

$\parallel v - \dfrac{v}{\| v\|} \parallel \leq 1-\sqrt{1-a-b} $

How is this proven?

Answer 1

The claim (as it was originally written -- the question has now morphed to something rather different) is perhaps not far from a correct one, but it's false as written.

Write $v = cu$ where $u$ has length $1$ (i.e., $u = v / \| v \|$) and $c = \|v\| \ge 0$. Then $$ \| v - \frac{v}{\|v\|}\| = \|cu - u\| = |c - 1| $$

Now since $$ a \le \|v\|^2$$ we get $$ a \le c^2 $$ hence (if $a$ is positive, which was not specified), $$ \sqrt{a} \le c\\ \sqrt{a} - 1 \le c - 1. $$

That's about as close to your claim as we can get, because the claim itself is false. For instance, in the reals, let $v = 1$. THen we have $$ -4 < |v|^2 < 12 $$ which should imply that $$ 0 \le 1 - \sqrt{-4}, $$ which is nonsense.

Assuming that you forgot to include that $a \ge 0$, the claim is still wrong, as you can see from the case

$$ \frac{1}{4} \le \frac{1}{2}^2 \le 1. $$ for in this case, the conclusion should be that $$ | \frac{1}{2} - 1 | < \frac{1}{4} - 1, $$ which is false.