Show that $\Vert x-y\Vert\ge\tfrac12\max(\Vert x\Vert,\,\Vert y\Vert)\left\Vert\tfrac{x}{\Vert x\Vert}-\tfrac{y}{\Vert y\Vert}\right\Vert$ Given that $x,\,y$ belong to a normed vector space with $0\ne x\ne y\ne0$.
i've tried taking $\Vert x \Vert$ as the max and tried to factor by the norm of x but it lead me to nowhere. i also tried using the inverse triangular inequality.
 A: WLOG we may assume $\|x\|\ge\|y\|$. So we need to prove
$$\left\lVert x-\frac{\|x\|}{\|y\|} y\right\rVert\le2\lVert x-y\rVert.$$
Using the triangle inequality, we get
$$\left\lVert x-\frac{\|x\|}{\|y\|} y\right\rVert=\left\lVert (x-y)+\left(y-\frac{\|x\|}{\|y\|} y\right)\right\rVert\le\lVert x-y\rVert+\left(\frac{\lVert x\rVert}{\lVert y\rVert}-1\right)\lVert y\rVert.$$
The right-hand side equals
$$\lVert x-y\rVert+\lVert x\rVert - \lVert y\rVert.$$
Using the triangle inequality again yields
$$\lVert x \rVert - \lVert y\rVert = \lVert (x-y)+y\rVert - \lVert y\rVert \le\lVert x-y\rVert+0,$$
which proves our initial assertion.
A: Without loss of generality assume $\|x\|=\max\{\|x\|,\|y\|\}$. Then your problem is equivalent to shown whether the following holds:
$$
 \Big\| \frac{x}{\|x\|}-\frac{y}{\|x\|}\Big\|\ge\frac12\left\Vert\frac{x}{\Vert x\Vert}-\frac{y}{\Vert y\Vert}\right\|
$$
Let $u_x=\frac{x}{\|x\|}$ and $u_y=\frac{y}{\|y\|}$
Then by the triangle inequality
\begin{align}
\Big\|u_x-\frac{\|y\|}{\|x\|}u_y\Big\|&\geq \|u_x-u_y\|-\Big(1-\frac{\|y\|}{\|x\|}\Big)\\
&\geq \|u_x-u_y\| - \Big\|u_x-\frac{\|y\|}{\|x\|}u_y\Big\|
\end{align}
since $1-\frac{\|y\|}{\|x\|}\leq \Big\|u_x-\frac{\|y\|}{\|x\|}u_y\Big\|$
