Here's another proof. 
By taking sub-sequences and rescaling we may assume that $a_n=0$, $c_n\geq 1$ and $F(0)=0$.
Define a function $G\colon [1,\infty)\times (1,\infty)\to \mathbb R$ by
$$
G(x,y)=\tfrac 1 y F(xy)-F(x).
$$
**Claim.** $G$ is positive and increasing in both $x$ and $y$.

**Proof.**
Since $F$ is strictly convex, $F(0)=0$,  and $x = 1/y(xy)+(1-1/y)0$, $G(xy)>0$. Since $F'$ is increasing, we have $\frac {d}{dx}G(xy)=F'(xy)-F'(x)>0$, so $G$ increases in $x$. Since $F'$ is increasing and $G(x,y)=1/y\int_0^yF'(xt)x\,dt - F(x)$, $G$ increases in $y$, completing the proof of the claim.

Suppose there is $\lambda_0>0$ such that $\lambda_n\geq \lambda_0$ for all $n$. We must show that $D_n\not\to 0$. Indeed, let $y_0=1/(1-\lambda_0)>1$,
$$
D_n = G(c_n,1/(1-\lambda_n))\geq G(1,y_0)>0.
$$
QED