proof of Hilbert space is uniformly convex space

We prove that in fact an inner product space is uniformly convex.Let $ϵ>0$, $u,v\in H$ such that $\parallel u\parallel \le 1$, $\parallel v\parallel \le 1$, $\parallel u-v\parallel \ge ϵ$. Put $\delta =1-\frac{1}{2}\sqrt{4-{ϵ}^2}$.Then $\delta >0$ and by the parallelogram law

${\parallel u+v\parallel }^2$$=$${\parallel u+v\parallel }^2+{\parallel u-v\parallel }^2-{\parallel u-v\parallel }^2$

$=$$2{\parallel u\parallel }^2+2{\parallel v\parallel }^2-{\parallel u-v\parallel }^2$

$\le$$4-{ϵ}^2$

$=$$4{(1-\delta )}^2.$

Hence, $\parallel \frac{u+v}{2}\parallel \le 1-\delta$.

Since a Hilbert space is an inner product space,a Hilbert space the conditions of a uniformly convex space.