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:

Assume by contradiction that $\lambda_n \not\to 0$. Then by passing to subsequences we can assume that $ a_n \to a , \lambda_n \to \lambda \in (0,1]$. Recall that $b_n,c_n \to \infty$. We replace $(1)$ by $ \lambda_n a +(1-\lambda_n) b_n=\tilde c_n$.

Define $\tilde D_n=\lambda_n F(a)+(1-\lambda_n)F( b_n)-F\big(\tilde c_n\big) $.

We want to prove that $D_n \to 0$ implies $\tilde D_n \to 0$.

$D_n-\tilde D_n=\lambda_n \big(F(a_n)-F(a)\big)+\big(F(\tilde c_n)-F(c_n)\big)$. The first term tends to zero, since $F(a_n) \to F(a)$.

The problem is that I am not sure how to prove that $F(\tilde c_n)-F(c_n) \to 0$, since $c_n,\tilde c_n$ are unbounded. If $M:=\sup \{ F'(x) \, | \, x \in [0,\infty) \}< \infty$, then $|F(\tilde c_n)-F(c_n)| \le M|c_n-\tilde c_n| \to 0$. But what happens when $F'(x) \to \infty$ when $x \to \infty$?