We have $(A^{-1})^T = (A^T)^{-1}$ for any invertible matrix. It follows from this that if $A$ is invertible and symmetric $$(A^{-1})^T = (A^T)^{-1} = A^{-1}$$ so $A^{-1}$ is also symmetric. Further, if all eigenvalues of $A$ are positive, then $A^{-1}$ exists and all eigenvalues of $A^{-1}$ are positive since they are the reciprocals of the eigenvalues of $A$. Thus $A^{-1}$ is positive definite when $A$ is positive definite.
As already pointed out, this is not true for a general matrix norm. We can develop an elementary proof and see what assumptions come up. We will show, using the hint, that if $A+B$ is not SPD, then $\|A^{-1}\|\|B\|\geq 1$.
Assume that $A+B$ is not positive definite, that is, it has at least one non-positive eigenvalue. Let $f(t):=A+tB$ where $t$ is a real scalar. Note that $A=f(0)$ is positive definite and $A+B=f(1)$ is not. Since the eigenvalues of $f$ are continuous functions of $t$, there is a $t_*\in(0,1]$ such that $f(t_*)$ has a zero eigenvalue.
Let $\|\cdot\|_😈$ be a vector norm.
There exists an $x$ such that $\|x\|_😈=1$ and $(A+t_*B)x=0$, so $x=-t_*A^{-1}Bx$ and
$$
1=\|x\|_😈=|t_*|\|A^{-1}Bx\|_😈\leq\|A^{-1}Bx\|_😈.
$$
Now if a matrix norm $\|\cdot\|_😃$ is consistent with the vector norm $\|\cdot\|_😈$, we have that
$$
1\leq\|A^{-1}B\|_😃.
$$
This means that if $\|A^{-1}B\|_😃<1$, then $A+B$ is SPD.
If you add sub-multiplicativity to your assumptions on the matrix norm, a sufficient condition is that $$\|A^{-1}\|_😃\|B\|_😃<1.$$
Best Answer
Here's a way to show that this condition holds directly. Let $e_1,\dots,e_n$ denote the standard basis (so that $e_i$ is the $i$th column of the identity matrix). By definition, it holds that $x^TBx \geq 0$ for any vector $x \in \Bbb R^n$.
Now, take $x = t e_i + t^{-1} e_j$ (for $t \neq 0$) and write out the expression $x^TBx$ as a function of $t$. Using the fact that $x^TBx$ is positive for all choices of $t \in \Bbb R\setminus \{0\}$, reach the desired conclusion about the entries $b_{ii},b_{ij},b_{ji},b_{jj}$.