Let $F:[0,\infty) \to [0,\infty)$ be a $C^1$ strictly convex function.

Let $\lambda_n \in [0,1],a_n\le c<b_n \in [0,\infty)$ satisfy $$ \lambda_n a_n +(1-\lambda_n)b_n=c_n \tag{1}$$ and assume that $c_n \to ֿ\infty$. (which implies $b_n \to ֿ\infty$). $c>0$ is just some constant, to make $a_n$ bounded.

Set $D_n=\lambda_nF(a_n)+(1-\lambda_n)F(b_n)-F\big(c_n\big) $, and assume that $\lim_{n \to \infty}D_n=0$

Question: Does $\lambda_n \to 0$?

My intuition is that even if $F$ becomes "less convex" (closer to being affine) when $x \to \infty$, then we cannot put to much weight on the $a_n$-since otherwise we get hit by the "convexity gap" between $a_n$ and $b_n$ by a non-negligible amount, which should make $D_n$ large.

Edit:

This is an attempt to understand Ron P's answer:

We have $D(a_n,c_n,b_n)=\lambda_n F(a_n)+(1-\lambda_n)F( b_n)-F(c_n)$, where $ \lambda_n a_n +(1-\lambda_n) b_n=c_n$.

Similarly, $D(a,c_n,b_n)=\tilde \lambda_n F(a)+(1-\tilde \lambda_n)F( b_n)-F(c_n)$, where $ \tilde\lambda_n a +(1-\tilde \lambda_n) b_n=\tilde c_n$.

Suppose that $a_n \to a$. (This implies $\lambda_n-\tilde \lambda_n \to 0$). We have

$$D(a_n,c_n,b_n)-D(a,c_n,b_n)=\lambda_n F(a_n)-\tilde \lambda_n F(a)+(\tilde \lambda_n-\lambda_n)F(b_n). \tag{2}$$ The first term tends to zero, since $F(a_n) \to F(a)$ and $\lambda_n-\tilde \lambda_n \to 0$.

Why does the second term tend to zero? we don't have control over $F(b_n)$, right?