Let $v,w \in \mathbb{R}^n$ satisfy $v^Tw>0$.
Does $v^TBw \ge 0$ for every symmetric positive-definite matrix $B$?
I think the answer is negative; this is connected to the fact that the product of symmetric positive-definite matrices is not necessarily positive-definite:
Here is an example where $v^TBw =0$. The question is whether this $v^TBw$ could be strictly negative.
Set $$A= \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} \quad \text{ and } \quad B = \begin{bmatrix} 5 & 2 \\ 2 & 1 \end{bmatrix}.$$ Then $$AB = \begin{bmatrix} 12 & 5 \\ 7 & 3 \end{bmatrix}.$$
$A,B$ are symmetric positive-definite.
$$AB+(AB)^T= \begin{bmatrix} 24 & 12 \\ 12 & 6 \end{bmatrix}$$ which is singular. Thus, there exists a vector $v \in \mathbb{R}^2$ such that $$ \big(AB+(AB)^T \big)w=0. $$ Set $v:=Aw$. Then $$ v^Tw=w^TAw>0, $$ but $$ v^TBw=w^TABw=\frac{1}{2} w^T\big(AB+(AB)^T \big)w=0. $$
As Exodd commented, the answer is negative: Take for example $$ v=(−1,1) , w=(1,2), B=\text{diag}(100,1). $$
More generally, since positive definite matrices are orthogonally diagonalizable, the problem is reducible or equivalent to considering positive diagonal matrices $B=D$:
In that case the assumption is $$ v^Tw=\sum_i v_iw_i>0. $$ and the question is whether or not $$ v^TDw=\sum_i \lambda_iv_iw_i>0, $$ for any positive $\lambda_i$.
Thus it is clear that if there is an index $i$ such that $v_iw_i<0$, one can magnify its impact by choosing very large $\lambda_{i}$ to make the inner product negative.