Let $H$ be a Hlbert space. If $T$ is a bounded linear operator on $H$, then $|T|=\sqrt{T^*T}$ is called the absolute value of $T$. And if $A$ and $B$ are self-adjoint bounded linear operators, then we say that $A\leq B$ if $B-A$ is a positive operator. Now show that for any bounded linear operators $S$ and $T$, we have $|S+T|^2\leq2|S|^2+2|T|^2$.
I'm not sure how to proceed. I tried expanding out $|S+T|^2$ using the definition of absolute value, but it didn't lead anywhere productive. Note that I'd like to do this without referencing either Borel functional calculus or spectral theory.
It is a useful fact to know that there is a parallelogram law for operators in $\mathcal B(H)$. Namely, by expanding you may verify the following equality for $S,T \in \mathcal B(H)$ : $$ (S+T)^*(S+T) + (S-T)^*(S-T) = 2(S^*S + T^*T) $$
Note that $|V|^2 = V^*V$ for any operator $V$. This allows us to rewrite the above as $$2(|S|^2 + |T|^2) - |S+T|^2 = |S-T|^2$$
which is a positive operator for fairly obvious reasons, giving the desired inequality.
Further details may be found in Section 3.4 of Analysis Now, by Gert K. Pedersen.