Positive operator T is invertible iff  is strictly positive 
To prove that invertibilty implies the inner product is strictly positive.
I am trying to prove it by showing that Tv cannot be orthogonal to v, but can't seem to go anywhere. Any suggestion?
 A: If $T$ is positive then the following Cauchy-Schwarz inequality holds
$$|\langle Tu,v\rangle |^2\le \langle Tu,u\rangle \langle Tv,v\rangle \quad (*)$$ Assume $\langle Tv,v\rangle =0$ for some $v\neq 0.$ Then substituting $u=Tv$ in $(*)$ implies $Tv=0,$ hence $T$ is not injective.
Conversely assume $T$ is not injective. Then $Tv=0$ for some $v\neq 0.$ In particular $\langle Tv,v\rangle =0.$
When the space $V$ is finite dimensional the injectivity is equivalent to invertibility. For infinite dimension injectivity does not imply invertibility. Therefore a stronger assumption is needed, namely $$\langle Tv,v\rangle \ge c\|v\|^2$$ for a positive constant $c.$
A: You can prove this by using the fact that any positive operator has a unique positive square root.
Let's say that the unique square root is $R$, so $T= R^2=R^*R$ (since $R$ is also positive $R^*=R$.
Thus, $$\forall v\neq 0, \  \langle Tv,v\rangle=\langle Rv,Rv\rangle=\|Rv\|^2>0$$
This implies that $$ v= 0 \iff Rv = 0$$   so $R$ is injective, so $R$ is invertible, and since $$ T=R^2$$
$T$ is indeed invertible.
Even though this proof only proves if T is invertible then $\langle Tv,v\rangle >0$, you can use $T = R^*R$ this trick to solve both directions
