Showing that these inner product induced norms are equivalent

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I am working on the following exercise:

Let $H$ be a Hilbert space, $T:H\to H$ be a self-adjoint isomorphism and suppose that $T$ is positive, that is, $\langle Tx,x\rangle\geq 0$ for all $x\in H $. Define $$[x,y]= \langle Tx,y\rangle, x,y\in H,$$ and show that this is an inner product in $H$ an that the norm induced by this inner product and the original norm of $H$ are equivalent.

Let $\|\cdot\|_0$ denote the norm induced by $[\cdot,\cdot]$ and $\|\cdot\|$ denote the original norm of H. I'm struggling to prove that $[x,x]>0$ for all $x\in H\backslash\{0\}$ and $\|\cdot\|\leq C\|\cdot \|_{0}$ for some constant $C>0$. I've done all the rest.

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We have

$$\lVert x\rVert^2 = \langle x,x\rangle = [T^{-1}x,x] \leqslant \lVert T^{-1}x\rVert_0\lVert x\rVert_0 \leqslant K\cdot \lVert T^{-1}x\rVert\cdot \lVert x\rVert_0 \leqslant K\cdot \lVert T^{-1}\rVert_{\text{op}} \lVert x\rVert\cdot\lVert x\rVert_0$$

by the Cauchy-Schwarz inequality for $[\,\cdot\,,\,\cdot\,]$, the inequality $\lVert y\rVert_0 \leqslant K\lVert y\rVert$, and the definition of the operator norm (with respect to $\lVert\,\cdot\,\rVert$). For $x\neq 0$, we can divide by $\lVert x\rVert$ and obtain

$$\lVert x\rVert \leqslant K\cdot \lVert T^{-1}\rVert\cdot \lVert x\rVert_0\,.$$

This inequality is of course also true for $x = 0$.