Assume $0<c_i<1$ and $d_i>0$ for $i=1,\dots k$ such that $\sum_{i=1}^kc_i=1$ and $\sum_{i=1}^kc_id_i=1$.
I am wondering if $$\sum_{i=1}^kc_id_i^3\ge \sum_{i=1}^kc_id_i^2?$$
I tried some examples and this inequality holds, but I don't know if it can be proved.
On
Here is a proof using convex functions: For each $d > 0$ is $x \mapsto d^x = e^{x \log d}$ a convex function. Consequently, $$ f(x) = \sum_{i=1}^k c_i d_i^x $$ is convex for $x \in \Bbb R$. In particular, $f(n+1) - f(n)$ is increasing in $n$, so that $$ \sum_{i=1}^kc_id_i^3 - \sum_{i=1}^kc_id_i^2 \ge \sum_{i=1}^kc_id_i - \sum_{i=1}^kc_i = 1 - 1 = 0 \, . $$
More generally, $f(0) = f(1)$ implies that $f(x)$ is increasing for $x \ge 1$, so that $$ \sum_{i=1}^kc_id_i \le \sum_{i=1}^kc_id_i^2 \le \sum_{i=1}^kc_id_i^3 \le \ldots $$
Note, if $d\ge1$, then $cd^2(d-1)\ge c(d-1)\ge0$ (given that $c\ge0$), while if $0\lt d\le1$ then $0\le(1-d)cd^2\le(1-d)c$, so the inequality $cd^2(d-1)\ge c(d-1)$ still holds. Thus
$$\sum cd^3-\sum cd^2=\sum cd^2(d-1)\ge\sum c(d-1)=\sum cd-\sum c=1-1=0$$