Suppose $A$ and $B$ belong to a kind of special hermitian matrices, which have such properties: $1.$ contain only one negative eigenvalue. $2.$ the negative eigenvalue and the second-largest positive eigenvalue are opposite to each other. $3.$ trace 1.

Let $C$=$\alpha A+(1-\alpha)B$, $\alpha \in [0,1]$. Then, we can find that the absolute value of the minimum eigenvalue of $C$ is always less than the second-largest positive eigenvalue of $C$ (just like an upper bound).

Can this observation be proved?

Suppose $A$ and $B$ belong to a kind of special hermitian matrices, which have the following properties:

1. $A$ and $B$ contain only one negative eigenvalue.
2. the negative eigenvalue and the second-largest positive eigenvalue are opposite to each other.
3. $\operatorname{trace}(A) = \operatorname{trace}(B)= 1$.

Let $$ C=\alpha A+(1-\alpha)B,\quad \alpha \in [0,1]. $$ Then, we can find that the absolute value of the minimum eigenvalue of $C$ is always less than the second-largest positive eigenvalue of $C$ (just like an upper bound).

Can this observation be proved?