Is there any constant $c$ satisfying the inequality

Question

Is there any positive constant $c\in \mathbb{R}$ such that for any nonzero $a,b\in \mathbb{R}^n$,

$$\frac{1}{2}\|a-b\|^2-\frac{1}{\pi}\sqrt{\|a\|^2\|b\|^2-\langle a,b\rangle^2}+\langle a,b\rangle\frac{\cos^{-1}(\rho)}{\pi}\geq c\|a-b\|^2$$

where $\rho=\frac{\langle a,b\rangle}{\|a\|\,\|b\|}$.

I checked extreme cases where $a,b$ are orthogonal or in the same or opposite directions. It seems that there is a positive $c$, satisfying the above inequality.

Answer 1

We have two vectors $a$, $b$ enclosing an angle $\phi\in[0,\pi]$, and are told to prove that $${1\over2}|a-b|^2-{1\over\pi}|a|\,|b|\sin\phi+|a|\,|b|\cos\phi\,{\phi\over\pi}\geq c|a-b|^2$$ for a universal $c>0$. This claim can be written as $$|a|\,|b|{\sin\phi-\phi\cos\phi\over\pi}\leq\left({1\over2}-c\right)|a-b|^2\ .$$ Now one can verify (by drawing the graphs) that $$0\leq{\sin\phi-\phi\cos\phi\over\pi}\leq{1\over2}(1-\cos\phi)\qquad(0\leq\phi\leq\pi)\ .$$ Therefore it is sufficient to prove $${1\over2}|a|\,|b|(1-\cos\phi)\leq\left({1\over2}-c\right)|a-b|^2\ .\tag{1}$$ Note that by the cosine theorem $$|a-b|^2=|a|^2+|b|^2-2|a|\,|b|\cos\phi=\bigl(|a|-|b|\bigr)^2+2|a|\,|b|(1-\cos\phi)\ .$$ This proves $${1\over2}|a|\,|b|(1-\cos\phi)\leq{1\over4}|a-b|^2\ ,$$ so that $(1)$ is true with $c={1\over4}$.

Answer 2

Your inequality is equivalent to find $c\ge 0$ such that for all $\theta $, $$ \|a\|\|b\|\cos(\theta)\theta\geq \|a\|\|b\|\sin(\theta)+{\pi}(c-0.5)||a-b||^2$$ ($\langle a,b\rangle=\|a\|\|b\|\cos(\theta)$) so the inequality should hold with $c=0$

$\|a-b\|^2=\|a\|^2+\|b\|^2-2\|a\|\|b\|\cos(\theta)$ replacing we would get $$\|a\|\|b\|(\sin(\theta)-\theta\cos(\theta)+\pi\cos(\theta))\le \frac{\pi}{2}(\|a\|^2+\|b\|^2)$$ so by AGM inequality for $\|a\|=\|b\|$ we require $\sin(\theta)-\theta\cos(\theta)+\pi\cos(\theta)-\pi\le 0$ which isn't true for all $\theta$.