Bounded linear operator with a "lower bound" in Hilbert space is invertable I am having a bit of difficulty with the following homework problem.

Assume $H$ is an Hilbert space, $A\colon H\to H$ is a bounded linear operator. There exists $C>0$ such that
$$C\| x\|^2\le\lvert \langle Ax, x\rangle\rvert , \quad \forall x\in H$$
Prove that $A$ is bijective.

I know how to prove $A$ is injective:

Assume two different elements $x, y\in H$ have same image under $A$, i.e., $A(x)=A(y)$. Consider $z=x-y$ which should not be zero according to our assumption. However, we can see contradiction that
$$ C\| z\|^2\le \lvert \langle Az, z\rangle\rvert =\lvert \langle 0, z\rangle\vert =0\Longrightarrow z=0$$

However, I don't know how to prove $A$ is surjective.
 A: Finally I find the solution. Proof of surjection could be done by the following 2 steps:

  
*
  
*Range of $A$, denoted by $R(A)$, is a closed set. 
  

Say $\{y_n\}$ is a converged set in $R(A)$, and $\{y_n\}\to y$. We know $R(A)$ is closed iff $y\in R(A)$ by properties of closed sets. By definition of $R(A)$, there exists a series $\{x_n\}\in H$ which satisfies $y_n=Ax_n, \forall n\in \mathbb N$. Then we have $\forall m,n\in \mathbb N$, 
$$\begin{align*} C\lVert x_m-x_n \rVert^2 &\le \lvert \langle Ax_m-Ax_n, x_m-x_n\rangle\rvert \\ &= \lvert \langle y_m-y_n, x_m-x_n\rangle\rvert \\  &\le \lVert y_m-y_n\rVert\cdot \lVert x_m-x_n\rVert \end{align*}$$
Cancel $\lVert x_m-x_n\rVert$ in both sides and we yields
$$\lVert x_m-x_n\rVert \le \frac{1}{C}\lVert y_m-y_n\rVert $$
which implies that $\{x_n\}$ is a Cauchy series in $H$. $H$ is Hilbert and thus Banach and thus complete, so $\{x_n\}$ converge to a certain element in $H$, say $x$. Every bounded linear operator is continuous in a Hilbert space, so we have
 $$y=\lim_{n\to\infty} y_n=\lim_{y\to\infty}Ax_n=A\lim_{n\to\infty}x_n=Ax\in R(A)$$


  
*$H=R(A)$
  

Because $R(A)$ is closed, we can decompose $H$ in the following way
 $$H=R(A)\oplus R(A)^\bot$$
where $R(A)^\bot$ is orthogonal complement of $R(A)$. We will prove $R(A)^\bot=\{0\}$ next. By definition of orthogonal complement, we have $\forall x\in R(A)^\bot$, $ \langle Ax, x\rangle=0$ because $Ax\in R(A)$. So we can derive 
$$ C\lVert x\rVert^2\le\lvert \langle Ax,x\rangle\rvert=0\Longrightarrow x=0 $$ which implies $R(A)^\bot=\{0\}$. So from direct sum decomposition of $H$ we know $H=R(A)$.
By now, we almost finished our proof. Just notice that $\forall y \in H$, $y\in R(A)$, so there exists $x\in H$ which satisfies $y=Ax$, which implies $A$ is surjective.
