If $A \in \mathcal{L}(H)$ and $\langle A(u),u \rangle \geq \langle u, u \rangle$, then $A$ is invertible. Exercise :

Let $H$ be a Hilbert space and $A \in \mathcal{L}(H)$ such that :
  $$\langle A(u),u \rangle \geq \langle u, u \rangle \; \forall u \in H$$
  Show that $A$ is invertible.

Attempt/Thoughts :
The inequality given to hold can be transformed to 
$$\langle A(u),u \rangle \geq \|u\|^2$$
since $\|u\| = \sqrt{\langle u, u \rangle} $ by the definition of the inner product functional for Hilbert spaces.
Now, by the Cauchy-Schwarz inequality, one can yield :
$$|\langle Au, u\rangle| \leq \|Au\|\|u\|$$
Combining the two expressions now gives us :
$$\langle Au, u \rangle \geq \|u\|^2 \Rightarrow |\langle Au,u\rangle| \geq \|u\|^2  \Rightarrow \|Au\| \|u\| \geq \|u\|^2$$
$$\implies$$
$$\boxed{\|Au\| \geq \|u\|}$$
Now if I consider a sequence $\{u_n\}_{n \in \mathbb N} \subset H$ such that $Au_n \to u \in H$ then it would be :
$$\|A(u_n - u_m) \| \to 0 \implies \|u_n-u_m\| \to 0 \quad \text{for} \; n,m \to \infty$$
That means that $\{u_n\}_{n \in \mathbb N}$ is Cauchy and thus $A : H \to A(H)$ is injective $("1-1")$, thus invertible ?
Is my approach correct ? Any tips, corrections and/or elaborations will be appreciated.

If $Au = Av$ then $$\|u-v\| \le \|Au - Av\| = 0$$ so that $u=v$ too.
