Symmetric Positive Definite Matrix Plus Symmetric Matrix is again Positive Definite Let $|| \cdot ||$ be a matrix norm (need not to be induced or submultiplicative), $A \in \mathbb{R}^{n \times n}$ a symmetric, positive definite, (therefore invertible) matrix and $B \in \mathbb{R}^{n \times n}$ be just a symmetric matrix.
Show that if
$$
||A^{-1}|| \cdot ||B|| < 1
$$
then $A + B$ is positive definite.
Hint: If $t \mapsto A(t) \in \mathbb{R}^{n \times n}$ is a continuous function, there are continuous functions which maps $t$ to each Eigenvalues of $A$.

I just don't have much of an idea where to start. Since the norm need not to be submultiplicative, I can't say $||A^{-1}|| \geq (||A||)^{-1}$. I don't know how to use the hint and using the diagonalization of the matrices doesn't seem to help.
EDIT: The things I wrote below, I don't think that even holds because you probably need again the submultiplicative property to conclude that.
Only thing I noticed was that $$
||A^{-1}|| \cdot ||B|| < 1 \\
\Rightarrow ||D_{A^{-1}}|| \cdot ||D_B|| < 1
$$
where $D_{A^{-1}}$ and $D_B$ are the diagonal form of the matrices with Eigenvalues as their entries, but since the norm is not induced, I don't know how this could help.
Thank you very much for your help.
 A: 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 $$\|A^{-1}\|_\|B\|_<1.$$
