I think you have \[A_{n} (m \times m)\]
Use
1. definition of convex function \[f(\alpha x+(1-\alpha)y)\leq \alpha f(x) + (1-\alpha ) f
\]
2. \[det(cA)=c^m det(A)\]
3. Show 1. using Minkowski's inequality \[A\neq 0, B\neq 0 (m\times m)\] and Positive semidefinite
\[[[det(A+B)]]^{\frac{1}{m}}\geq (detA)^{\frac{1}{m}} + (detB)^{\frac{1}{m}}\]
Then, you will have
\[\alpha f(x) = -[det(\alpha A_{0} + \alpha x_{1}A_{1}+\cdots +\alpha x_{n}A_{n})]^{\frac{1}{m}}\]
and the same for \[(1-\alpha ) f
\]
and
\[f(\alpha x + (1-\alpha )y) = -[det\{ (\alpha + (1-\alpha) )A_{0} + (\alpha x_{1}+(1-\alpha )y_{1} )A_{1}+\cdots +(\alpha x_{n} + (1-\alpha )y_{n})A_{n} \}]^{\frac{1}{m}}\]