Given \begin{align} f(x)=\sum_{k=0}^n { n \choose k} (1-x)^{n-k} \cdot x^k \cdot a_{n-k} \cdot b_k \end{align}
Find \begin{align} \min _{x \in [0,1] } f(x) \end{align}
We can assume that $a_k$ and $b_k$ are positive and monotone increasing sequences.
Note, that I am not interested in the minimizer $x^{*}$ but rather a minimum value of $f(x^{*})$.
EDIT
Based on the comments of A.S., we can redefine the problem:
Given
$$\begin{align} g(x)=\sum_{k=0}^n { n \choose k} (1-x)^{n-k} \cdot x^k \cdot c_k \end{align}$$
for some $c_k>0$, find
$$\begin{align} \min _{x \in [0,1] } g(x) \end{align}$$
This problem is related to the question I posted here.

The function $g$ is a linear combination of Bernstein polynomials. As these polynomials are a basis of the space of polynomials, your question is equivalent to asking for the minimum value of an arbitrary polynomial of degree $n$, which is impossible to answer in this general form.
For Bernstein polynomials you can obtain upper and lower bounds of the function values from the coefficients $c_k$: $$ \min_k c_k \le g(x) \le \max_k c_k \quad\forall x\in[0,1]. $$ The proof of this is simple: $$ g(x) =\sum_{k=0}^n {n \choose k}(1-x)^{n-k}x^kc_k \le \max_k c_k \cdot \sum_{k=0}^n {n \choose k}(1-x)^{n-k}x^k = \max_k c_k \cdot 1, $$ which also shows that the bound is sharp if the $c_k$'s are all of the same value.