For $\langle Tx, x \rangle \geq \|x\|^2$, prove a solution exists for $Tx = y$. Edit

I've posted this a couple other times, so I now plan on deleting those, and just using this one.

Here's the original problem:


  $H$ is a real Hilbert space. Let $T: H\longrightarrow H$ be a bounded linear operator on $H$ such that the inner product satisfies
  $$\langle Tx, x \rangle \geq \|x\|^2 \tag{A}$$
  Prove that a solution for the equation $Tx = y$ exists for all $y \in H$.


I've been looking at this problem and studying orthogonal complements and the direct sum theorem for a few days now, to no avail, so I suspect there's something very fundamental I don't understand here. Below I'm posting W. Zhan's proof that I don't understand, and a couple of starting points of my own. W. Zhan's, as mentioned below, isn't working for me, so I'm starting a bounty. Thank you.



Proof given by W. Zhan


  Note that adjoint of a bounded operator always exists by Riesz Represenation Theorem. So we have 
  $$ \langle Tx,x \rangle = \langle x, T^*x \rangle  \ge ||x||^2$$ 
  So $T^*$ is injective. We also have $im T^\perp =  \ker T^*$. Hence, $im T = H$.


I'm not sure what I'm not understanding. I do understand


  So $T^*$ is injective.


and also


  We also have $im T^\perp =  \ker T^*$


but not how those two facts yield the conclusion.



Here's a couple starts I've had on my own

Proof (alternate start 1)

Let $Tx = 0$, then 

$$0 = \langle Tx, x\rangle \geq \|x\|^2 \geq 0$$

so $\|x \| = 0$, but by definition of the norm, only if $x = 0$. So $Tx = 0 \iff x = 0$, therefore $T$ is injective. This means $T$ has an inverse linear operator $T^{-1}$.

Now, let $y \in \overline{TH}$. By definition, $y$ is a limit point of $TH$, so there exists a sequence $(x_n)$ in $H$ such that $(Tx_n) \longrightarrow y$. However, $x_n = T^{-1}Tx_n$, which implies

$$\lim_{n\to \infty}x_n = \lim_{n\to \infty} T^{-1}Tx_n  = T^{-1}\lim_{n \to \infty}Tx_n = T^{-1}y$$

... here I think I'm in trouble because I don't know if I can actually show $T^{-1}y$ exits, which I think is equivalent to showint $T^{-1}$ is bounded.

Proof (alternate start 2)

Here, as W. Zhan pointed out, it's easy to see that $(TH)^\perp = \{0\}$, which implies that $TH$ is dense in $H$. In other words, $\overline{TH} = H$. So all that remains is to show that $TH$ is closed. But this is similar to the argument in my first attempt.

Discussion

I understand that $T^*$ is injective—because it is a linear operator and $Tx = 0 \implies x = 0$. In fact $T$ is as well. So I believe another option is to show surjectivity of $T$, but I can see how to do that. 

Would somebody mind proving the original proposition in a detailed way?
Note that adjoint of a bounded operator always exists by Riesz Represenation Theorem. So we have
$$ \langle Tx,x \rangle = \langle x, T^*x \rangle  \ge ||x||^2$$
So $T^*x =0 \Rightarrow ||x||=0 $, hence $T^*$ is injective. We also have $im T^\perp =  \ker T^*$. Thus, $im T = H$.
To see why $T$ has closed range, note $||Tx||= \sup_{||y||\le1} |\langle Tx, y \rangle| \ge \langle Tx, x \rangle /||x|| \ge ||x||$.  Then result follows from characterization of closed image. The direction you need is actually already on MSE.
EDIT: Here's an easier way to see why $T$ has closed range. Suppose $Tx_n \rightarrow y$, then applying $x=x_n-x_m$ implies $x_n$ is Cauchy. Continuity implies $Tx = y$.
We can prove that $T$ is not singular, and from that, the problem is solved.

Assume $T$ is singular. Then there exists a vector $v \not= 0$ such that $Tv = 0$. But then $\langle Tv, v\rangle = 0 &lt; \|v\|^2$ which contradicts our assumption that $\langle Tv, v\rangle \ge \|v\|^2$.

Okay, so now that $T$ is not singular, it has an inverse, implying that $x = T^{-1} y$.