TITLE: Prove of inequality under a Hilbert space.
QUESTION [8 upvotes]: Let $x\neq y$ when $x,y\in H$ and H is a Hilbert space which satisfy $\|x\|=\|y\|=r$. Show that $\|\frac{x+y}{2}\|<r$. 
Actually in my question r=1 but as far as i could understand there is a way to prove this for any r. Is this true?
I tried to go with showing that $\|tx+(1-t)y\|<r$ where $t=\frac{1}{2}$ to no avail. Am I in a good direction? can someone point out the trick?

REPLY [3 votes]: So this is my final answer for any interested party answer and thank you @joy
$\|\frac{x+y}{2}\|^2 \leq \|\frac{x+y}{2}\|^2 + \|\frac{x-y}{2}\|^2 = \frac{1}{2}\|x\|^2+\frac{1}{2}\|y\|^2$
and since $x \neq y$:
$\|\frac{x+y}{2}\|^2 < \frac{1}{2}\|x\|^2+\frac{1}{2}\|y\|^2=\frac{1}{2}r^2+\frac{1}{2}r^2=r^2$
Hence:
$\|\frac{x+y}{2}\| < r$