Why we can use Cauchy-Schwartz inequality in Evans‘ proof on symmetric operator?

In Evans' book Partial differential equations, in Page 731,

Let $S:H\rightarrow H$ be a linear bounded symmetric operator on real Hilbert space. Write \begin{align} &m:=\inf_{||u||= 1}\langle Su,u\rangle &M=\sup_{||u||=1}{\langle Su,u\rangle} \end{align} Show the spectrum of $S$ : $\sigma(S)\subset [m,M]$ and $m,M\in \sigma(S)$.

I know how to show the first part. But here is a step of the proof of $M\in \sigma(S)$ in Evans: Since the pairing $[u,v]:=(Mu-Su,v)$ is symmetric and $[u,u]\ge 0$ for any $u\in H$, the Cauchy-Schwartz inequality implies $$|\langle Mu-Su,v\rangle|\le (Mu-Su,u)^{1/2}(Mv-Sv,v)^{1/2}$$ for any $u,v$ in $H$. I know if $[u,u]=0$ if and only if $u=0$ then we can safely apply the CS-inequality. But it seems in this case, we can not guarantee the $[u,u]=0$ implies $u=0$. So why can we apply the CS? Please note that if we don't have $[u,u]=0$ implies $u=0$, $[\cdot,\cdot]$ is not a inner product but only a semi-inner product? Can we use CS in this space?

You don't require the condition $[u,u]=0$ implies $u=0$ for C-S. Just use the fact that $[u+av,u+av] \geq o$ to get $[u,u]+|a|^{2}[v,v]+2 \Re a[v,u] \geq 0$. Let $a=-t[u,v]$ where $t \in \mathbb R$ and minimize over $t$ to get C-S. [ Or just put $t=\frac 1 {[v,v]}$. Note that if $[v,v]=0$ then we get $[u,u] -2t|[u,v]|^{2} \geq 0$ for all $t$ which implies $[u,v]=0$ so the proof works in this case also].
Cauchy-Schwarz holds for a pseudo inner product. But, in your case, you can also use a trick by adding $\epsilon\langle u,u\rangle$ to your pseudo inner product, and letting $\epsilon\downarrow 0$ in the following: $$|\langle (M-S+\epsilon I)u,v\rangle|^2 \le \langle (M-S+\epsilon I)u,u\rangle \langle (M-S+\epsilon I)v,v\rangle.$$