How to show that this bounded linear operator is onto? This is part of an old prelim exam in Analysis that I'm studying to prepare for my own prelim.
We're given that $H$ is a Hilbert space over $\mathbb R$ and $T\in\mathcal B(H,H)$ is a bounded linear operator such that
$$
\langle Tx, x\rangle \geq \lVert x\rVert^2\quad \forall x\in H
$$
The problem actually asks us to show that $Tx=y$ has a unique solution $\forall y\in H$ and satisfies $\lVert x\rVert \leq \lVert y\rVert$, but I got stuck just trying to show $T$ is onto. I don't really have any ideas for strategies here. I tried making something of the fact that the above implies $\langle Ty-y,y\rangle\geq 0\ \forall y$, but that didn't seem to go anywhere.

Due to the assumption, $T$ is injective.
Let $R(T)$ denote the range of $T$. Then $T^{-1}$ exists as a linear mapping from $R(T)$ to $H$. Due to the inequality, we have $\|T^{-1}y\|\le \|y\|$ for all $y\in H$.
Now argue $R(T)$ is closed: take a sequence $(y_k)$ in $R(T)$ converging to $y$ in $H$. Then by the estimate on $T^{-1}$ the sequence $(T^{-1}y_k)$ is Cauchy, thus converging to some $x$ with $Tx=y$. Hence, $R(T)$ is closed.
It remains to prove that $R(T)=H$. Take $x\in R(T)^\perp$. Then $\langle x, Ty\rangle=0$ for all $y$. Thus, $\langle x,Tx\rangle =0$ implying $x=0$.
So we have shown that $R(T)$ is closed, and its orthogonal complement is zero, this proves $R(T)=H$. 
