Suppose that we have some $z_k>0$, $k=\{0,1,\cdots,n\}$. I want to compare weighted averages of $z_k$'s when the weights are defined by binomial probabilities.
More specifically, for $p$ and $q$, where $p,q\in (0,1)$, and for some $\lambda\in(0,1)$, let $x=\lambda p+(1-\lambda)q$.
In this case, should the following be true for all $\lambda\in(0,1)$?
$$\max\bigg\{\sum_{k=0}^{n}{n\choose k}p^k(1-p)^{n-k}z_k,\sum_{k=0}^{n}{n\choose k}q^k(1-q)^{n-k}z_k\bigg\}\geq\sum_{k=0}^{n}{n\choose k}x^k(1-x)^{n-k}z_k.$$
I could see that it's true for $n=2$ but I can't show that it still holds for any $n>2$.
No, this is not the case, not even for $n=2$. You excluded $0$ and $1$, but by continuity we can still use them for a simple counterexample: For $p=0$, $q=1$, the left-hand side is zero whereas the right-hand side is positive for any $\lambda\in(0,1)$.