I tried to do this exercise but I'm not completely sure about my proof:
Let $(H, \langle\cdot ,\cdot \rangle)$ be a separable Hilbert space and let $A$ be a linear bounded operator from $H$ to itself s.t.
$$ \exists \epsilon >0 : \langle Ax ,x \rangle \ge \epsilon |x|^2 \quad \forall x \in H$$
Prove that:
(i) A is a bijection and its inverse is bounded
(ii) $|A^{-1}|_{\text{op}} \le \epsilon^{-1}$
(iii) if $A$ is compact and symmetric then $\sigma(A) \subset (0, |A|_{\text{op}}]$
Proof:
(i)
(i.i) Let $x \in H$ s.t. $|x| \ne 0$. By Cauchy-Schwarz we have $|Ax| ||x| \ge \epsilon|x|^2$ i.e. $|Ax| \ge |x|\epsilon$, then $|A|_{\text{op}} = \sup_{x : |x|=1} |Ax| \ge \epsilon $.
(i.ii) Suppose $x \in ker(A)$, then $Ax=0$ and $ 0=\langle Ax ,x \rangle \ge \epsilon|x|^2$ i.e. $x=0$, then $ker(A)= \{0\}$.
(i.iii) By contradiction suppose $Im(A) \ne H$, take $y \in (Im(A))^{\perp} \ne \emptyset$, then $0 = \langle Ay ,y \rangle \ge |y|^2$ i.e. $(Im(A))^{\perp} = \{0\}$, then $\overline{(Im(A))}=H$.
(i.iv) I want to show that $Im(A)$ is closed. Let $\{y_n\}_{n \in \mathbb{N}} \subset Im(A)$ s.t. $y_n \to y \in H$, I want to show $y \in Im(A)$ i.e. $\exists x \in H$ s.t. $y=Ax$.
$\{y_n\}_{n \in \mathbb{N}}$ is a converging sequence and then it is also a Cauchy sequence.
Since $\{y_n\}_{n \in \mathbb{N}} \subset Im(A)$ we have that exists $\{x_n\}_{n \in \mathbb{N}} \subset H$ s.t. $y_n = Ax_n \quad \forall n \in \mathbb{N}$
By (i.i) we have that also $\{x_n\}_{n \in \mathbb{N}}$ is a Cauchy sequence. Then $\exists x \in H$ s.t. $x_n \to x$. Since $A$ is continuous and by uniqueness of the limit we have $y=Ax$.
Then $Im(A)=X$ i.e. $A$ is a bijection.
By bounded-inverse theorem $A^{-1}$ is bounded too.
(ii)
Take $y \in H$ s.t. $|y|=1$ then $1=|y|= |A(A^{-1}y)| \ge \epsilon |A^{-1}y|$ then $|A^{-1}|_{\text{op}} = \sup_{y : |y|=1} |A^{-1}y| \le \epsilon^{-1} $.
(iii)
If $A$ is compact then $H$ is finite dimensional (but I don't know if this is useful). Since it is symmetric I know from spectral theorem that the bigger (in absolute value) eigenvalue of $A$ is $|A|_{\text{op}}$. Then it is enough to show that if we have an eigenvalue it is strictly positive. Let $(\lambda, u) \in \mathbb{R} \times H \setminus \{0\}$ s.t. $Au=\lambda u$, then
$ \langle Au ,u \rangle = \lambda |u|^2 \ge \epsilon |u|^2 \Rightarrow \lambda \ge \epsilon >0$
Thanks in advance.
Everything is correct, as far as I can tell. It feels kind of weird that you seem to be working on a real Hilbert space, but that's just me.
In (iii), you don't need the Spectral Theorem. If $Ax=\lambda x$ and $\|x|=1$, then $$ |\lambda|=\|\lambda x\|=\|Ax\|\leq\|A\|\,\|x\|=\|A\|. $$ After you prove that all eigenvalues are positive, you are done.
By the way, $\sigma(A)\subset(0,\|A\|)$ doesn't require compactness.