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 $R(T)=H$.