Strictly positive inner product for a pair of non-zero, positive operators. Let $ A,B $ be non-zero positive operators on a infinite-dimensional separable Hilbert space $(H , \langle \cdot, \cdot \rangle)$. I am required to prove that there exists $u' \in H$ such that 
\begin{alignat*}{2}
\langle Au' , u'\rangle >0 \ \ \text{and} \ \ \langle Bu', u' \rangle >0.
\end{alignat*}
I am quiet stuck with this problem. For a few things I have tried. It is obvious that there exists $v,w \in H$ such that 
\begin{alignat*}{2}
\langle Av, v \rangle > 0 \ \ \text{and} \ \ \langle Bw, w\rangle>0.
\end{alignat*}
And I have tried to calculate 
\begin{alignat*}{2}
\langle A(v + w), v +w \rangle
\end{alignat*}
for which I have to now show for instance $ \text{Re} \langle Av, w \rangle \geq 0 $. Alternatively I could try to directly find a positive operator $E$ such that 
\begin{alignat*}{2}
\langle Eu \ , \ u\rangle \leq \langle Au, u \rangle \ \ \text{and} \ \ \langle Eu , u \rangle \leq \langle Bu, u \rangle.
\end{alignat*}
I have also tried to apply orthogonal projections and polarisations, etc, but to no success. Hopefully it's some trivial details which I have missed.
The spectral theorem for bounded self-adjoint operator is not at my disposal.
Could anyone provide me with some hint? Thanks!

Let $v,w\in H$ be as you have defined them.  For $t\in[0,1]$ put $x_t=tu+(1-t)w$, and define $f,g:[0,1]\to [0,\infty)$ by 
$$f(t)=\langle Ax_t,x_t\rangle,\quad g(t)=\langle Bx_t,x_t\rangle.$$
Note that $f$ and $g$ are non-zero polynomials (of degree at most $2$).  Argue that there is some point $t_0\in[0,1]$ such that both $f(t_0)>0$ and $g(t_0)>0$, which proves the result.
