Let $A \in \mathbb{R}^{m\times n}, D \in \mathbb{R}^{m\times p}, b \in \mathbb{R}^{m} $ and $c \in \mathbb{R}^{n}$. Define $$P =\left \{ x=(x_1,x_2,...,x_n) \in \mathbb{R}^{n} : Ax=b , \forall_i \text{ } x_i \geq 0 \right \}$$ Suppose that $P$ is not empty and limited (bounded). Define a function $$f(\lambda) = max\left \{ (c + D\lambda)^{t}x : x \in P \right \}$$ I would like to show that $f((1-\alpha)\lambda^1 + \alpha\lambda^2) \geq (1-\alpha)f(\lambda^1) + \alpha.f(\alpha^2), $ for every $\alpha \in \left [ 0,1 \right ], \lambda^1, \lambda^2 \in \mathbb{R}^p$.
I don't know how can do this, because $f$ is a function of $\lambda$. I would appreciate any kind of help.
Proof idea
Define $F_x({\lambda}) = (c + D\lambda)^T. $ where $ D \in \mathbb{R}^p$
Given $\lambda_1$ and $\lambda_2$ $\in$ $\mathbb{R}^p$ and $\alpha \in (0,1)$ $F_x((1-\alpha)\lambda_1 + \alpha\lambda_2)$ =
$(c + D(1-\alpha)\lambda_1 + \alpha\lambda_2)^{T}x)$
= $(c^T + [D(1-\alpha)\lambda_1]^T + [D\alpha\lambda_2)^{T}])x$
= $(c^{T}x + (1-\alpha)\lambda_1^{T}D^{T}x + \alpha \lambda_{2}^{T}D^{T}x $
= $(c^Tx + (1-\alpha)\lambda_1^{T}D^{T}x) + (c^Tx + \alpha \lambda_{2}^{T}D^{T}x)$
...
= $F_x((1-\alpha)\lambda_1) + F_x(\alpha\lambda_2)$
= $(1-\alpha)F_x(\lambda_1) + \alpha F_x(\lambda_2)$
I need help with some steps.