Q:

Let $T$ be a Hermitian operator and $\|v\|=1$, prove $\langle T^2(v),v\rangle \ge\langle T(v),v\rangle^2$

Let $T$ be a Hermitian operator on an inner product space $V$ over the field of complex numbers, $C$ (of a finite dimension), also assume $\|v\|=1$. Prove:
$$\langle T^2(v),v\rangle \ge\langle T(v),v\rangle^2$$

A:

It's not true as stated, because it has the wrong scaling behaviour with respect to $v$.
Were you supposed to assume $\|v\|=1$?
EDIT: With that assumption, the hint is overkill.  The result follows directly from Cauchy-Schwarz.  Note that $\langle T^2 v, v \rangle = \langle T v, T v \rangle$ since $T$ is Hermitian.

A:

Edit: As pointed out by Robert Israel, this follows immediately from Cauchy-Schwarz. I so wanted to use your (wrong and now deleted hint), that I actually basically reproved Cauchy-Schwarz... 
Develop the following
$$
((T-\alpha I)^2v,v)=\alpha^2\|v\|^2-2\alpha(Tv,v)+(T^2v,v).
$$
Note that for every $\alpha\in \mathbb{R}$, the operator $(T-\alpha I)^2=(T-\alpha I)^*(T-\alpha I) $ is hermitian positive so
$$
((T-\alpha I)^2v,v)\geq 0.
$$
We have a quadratic in $\alpha$ which is always nonnegative.
This means that its discriminant is nonpositive:
$$
\Delta=4((Tv,v)^2-(T^2v,v)\|v\|^2)\leq 0.
$$
Hence
$$
(Tv,v)^2\leq (T^2v,v)\|v\|^2.
$$
So your inequality holds for all $\|v\|\leq 1$.

