how to get the following inequality $\|S_nx − S_mx\|_{H}^2 \leq 2M(S_mx −S_nx, x)$? 
Let $H$ a Hilbert space and $(S_n)$ be a sequence of self-adjoint operators in $\mathcal{L}(H)$ such that $(S_{n}x- S_{n+1}x,x)\geq 0\,\,\, \forall\,\,\,n$. Assume that $\|S_n\| \leq M\,\,\, \forall\,\,\,n$, for some constant $M$. How do prove that
$$\|S_nx − S_mx\|_{H}^2 \leq 2M(S_mx −S_nx, x)\,.$$

Attempt:
I will use that $$0 \leq (T u, u) \leq\|u\|_H^2 \,\,\,\,∀u \in H$$ is equivalent to $$(T u, u) \geq \|T u\|_H^2 ~~~\forall~ u ∈ H$$ for $T$ linear operator self-adjoint. So choosing $Tx= S_{n}x- S_{n+1}x$, we use equivalence and we get $$(S_{n}x- S_{n+1}x, x) \geq \|S_{n}x- S_{n+1}x\|_H^2,$$
the problem appears now: using triangular inequality, if $n>m.$
\begin{align}
\|S_nx − S_mx\|_{H}^2 &\leq ( \|S_nx − S_{n-1}x\|_{H}+ \ldots+ \|S_{m+1}x - S_mx \|_{H})^2\\
& =   \|S_nx − S_{n-1}x\|_{H}^2+ \ldots+ \|S_{m+1}x - S_mx \|_{H}^2 \\&\phantom{abcde}+ 2 \sum_{0 \leq i<j\leq n} \|S_ix − S_{i-1}x\|_{H}\|S_jx − S_{j-1}x\|_{H}\\
&\leq (S_{n}x- S_{m}x, x)+ 2 \sum_{0 \leq i<j\leq n} \|S_ix − S_{i-1}x\|_{H}\|S_jx − S_{j-1}x\|_{H}.
\end{align}
I'm not seeing any other way to use the hypotheses, and I can't see where to put this $ M $, does anyone know how to do it?
 A: The operators $S_n-S_{m}$ are positive by assumption, when $m\geq n$. This implies that they admit a positive square root, which lets you do this: writing $T=S_n-S_{m}$,
$$
\|Tx\|^2=\|T^{1/2}T^{1/2}x\|^2\leq\|T^{1/2}\|^2\,\|T^{1/2}x\|^2=\|T\|\,\langle Tx,x\rangle.  
$$
And
$$
\|T\|=\|S_n-S_{m}\|\leq\|S_n\|+\|S_{m}\|\leq 2M. 
$$
Thus
$$
\|(S_n-S_{n+1})x\|^2\leq 2M\langle(S_n-S_{n+1})x,x\rangle. 
$$

Another way, without using square roots, in outlined by Brezis: in Problem 39 it says to use Exercise 6.24, which in turn says to use Proposition 6.9. The idea in Proposition 6.9 is to use convenient positive quadratic forms.
Assume first that $\|T\|=1$. Then
$$
\langle T^3x,x\rangle=\langle T\,Tx,Tx\rangle\leq\|T\|\,\|Tx\|^2=\langle T^2x,x\rangle
$$
As $T\geq0$, the form $[x,y]=\langle Tx,y\rangle$ is positive, so it satisfies Cauchy-Schwarz. Then
\begin{align}
\langle T^2x,x\rangle&=[Tx,x]\leq[Tx,Tx]^{1/2}[x,x]^{1/2}\\[0.3cm]
&=\langle T^3x,x\rangle^{1/2}\langle Tx,x\rangle^{1/2}\\[0.3cm]
&\leq\langle T^2x,x\rangle^{1/2}\langle Tx,x\rangle^{1/2},\\[0.3cm]
\end{align}
so $\langle T^2x,x\rangle^{1/2}\leq\langle Tx,x\rangle^{1/2}$ and then
$$\tag1
\langle T^2x,x\rangle\leq\langle Tx,x\rangle. 
$$
Now for arbitrary $T$, use $T/\|T\|$ in $(1)$  to obtain
$$
\langle T^2x,x\rangle\leq\|T\|\,\langle Tx,x\rangle. 
$$
