Consider a matrix $C$ of the form:
$$ C = \begin{pmatrix} A & \mathbb{1} \\ \mathbb{1} & B \end{pmatrix} $$
where all the blocks ($A,B$ and the identity matrices) are square and of the same size, and the matrices $A,B$ are symmetric and positive definite.
For context, I encountered a matrix of this form as the Hessian of a certain function for which I am trying to determine the conditions for a critical point to be a minimum.
Suppose I can find the eigenvalues of $A$ and $B$ separately. Can I exploit this knowledge to conclude that $C$ is posdef in some conditions?
Take $Z = \begin{pmatrix}X \\ Y \end{pmatrix}$
$$Z^* C Z = X^* A X + Y^*BY + X^*Y + Y^*X$$
But Cauchy-Schwarz gives us $$|X^*Y + Y^*X| = \left| \begin{pmatrix}Y \\ X \end{pmatrix}^* \begin{pmatrix}X \\ Y \end{pmatrix}\right| \leqslant \left\| \begin{pmatrix}Y \\ X \end{pmatrix} \right\| \left\| \begin{pmatrix}X \\ Y \end{pmatrix} \right\| = \|Z\|^2$$
And $$\|Z\|^2 = \|X\|^2 + \|Y\|^2$$ so $$Z^* C Z \geq X^* A X + Y^*BY - \left( \|X\|^2 + \|Y\|^2 \right)$$
If you ensure $\forall \lambda \in \operatorname{Spec} (A) \cup \operatorname{Spec} (B)$, $\lambda >1$, you get $$Z^* C Z >0$$
So you have $C$ definite positive hermitian matrix.
The condition on the spectrum is not necessary. An example was given in the comments. We could maybe refine the proof to get the condition: $$\min \operatorname{Spec} (A) \cdot \operatorname{Spec} (B) >1$$
Refinement Replace $X$ by $X'=\sqrt{A}^{-1}X$ and $Y$ by $Y'=\sqrt{B}^{-1}Y$ in the previous identity (it is equivalent), you get $$Z'^*CZ' = X^*X + Y^*Y + X^* \sqrt{A}^{-1} \sqrt{B}^{-1} Y + Y^* \sqrt{B}^{-1} \sqrt{A}^{-1} X$$
But since they are complex numbers, $$\left(X^* \sqrt{A}^{-1} \sqrt{B}^{-1} Y\right)^* = \overline{X^* \sqrt{A}^{-1} \sqrt{B}^{-1} Y} = Y^*\sqrt{B}^{-1} \sqrt{A}^{-1} X$$
So $$f(Z) = Z'^*CZ' = X^*X + Y^*Y + 2\Re \left(Y^* \sqrt{AB}^{-1} X\right) $$
Now you can study this to find the minimal value. Its gradient is $$df_Z (H) = \begin{pmatrix} 2\Re(X^*H + Y^* \sqrt{AB}^{-1} H) \\ 2\Re(Y^*H + X^* \sqrt{BA}^{-1} H) \end{pmatrix}$$
$df_Z = 0 \iff X = -\sqrt{AB}Y \text{ and } Y =- \sqrt{BA} X $
If we continue the study, it will be equivalent for $f(Z)>0$ if $Z \neq 0$ and for $\sqrt{A}B\sqrt{A}$ to have eigenvalues $>1$, and equivalently for $AB$ to have eigenvalues $>1$.