Stricty convex norm defined by means of the strictly convex norm of $\ell_2.$ 
Let $X$ be a normed space and $T:X\rightarrow \ell_2$ be an injective, 
  linear and bounded operator. Prove that the norm
  $$\||x\||:=\sqrt{\|x\|^2+\|Tx\|^2}$$ on $X$ is strictly convex: for $x,~y\in X,~x\neq y$ such that $\||x\||=\||y\||=1$, we have $$\||(1/2)x+(1/2)y\||<1.$$

Attempt. Suppose that for some $x,~y\in X,~x\neq y$ such that $\||x\||=\||y\||=1$ we have $\||(1/2)x+(1/2)y\||=1$, so
$$4=\||x+y\||^2=\|x+y\|^2+\|T(x+y)\|^2,$$
where $$\|T(x+y)\|^2=\|Tx+Ty\|^2=2\|Tx\|^2+2\|Ty\|^2-\|Tx-Ty\|^2$$
by linearity and parallelogram law. Also $\|x+y\|^2\leq \|x\|^2+\|y\|^2+2\|x\|\|y\|$, but i don t seem to be getting $\|Tx-Ty\|^2=0$, which would give me the contradiction $x=y.$
Thanks in advance for the help.
 A: Continuing your proof, you have shown
$$4 = \|x+y\|^2 + 2\|Tx\|^2 + 2\|Ty\|^2 - \|Tx-Ty\|^2 < \|x+y\|^2 + 2\|Tx\|^2 + 2\|Ty\|^2$$
since $x\neq y$ and $T$ is injective. 
On the other hand, the assumption on $x$ and $y$ give $\|x\|^2 + \|Tx\|^2 =1 $ and $\|y\|^2 + \|Ty\|^2 =1$. Now multiply the last two identities by $2$ and add to obtain 
$$4 = 2\|x\|^2+2\|y\|^2+2\|Tx\|^2+2\|Ty\|^2.$$
Altogether, 
$$\|x+y\|^2 > 2\|x\|^2+2\|y\|^2$$
which is absurd because $\|x+y\|^2 \leq (\|x\|+\|y\|)^2 = \|x\|^2+\|y\|^2+2\|x\|\|y\| \leq 2\|x\|^2+2\|y\|^2$.
A: $\newcommand{\vertiii}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 
    \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}$
If $X$ is also a Hilbert space, notice that
$$\vertiii x^2 = \langle x,x\rangle + \langle Tx, Tx\rangle = \langle (I + T^*T)x, x\rangle$$
where $T^* : \ell^2 \to X$ is the adjoint of $T$.
Now, for $x \ne y$ using the pararellogram identity for the quasi-inner product $\langle (I + T^*T)\cdot, \cdot\rangle$, we get
$$\vertiii {\frac12(x+y)}^2 = \frac12 \underbrace{\vertiii{x}^2}_{=1} + \frac12 \underbrace{\vertiii{y}^2}_{=1} - \frac14 \underbrace{\vertiii{x-y}^2}_{> 0} < 1$$
