Let $F:[0,\infty) \to [0,\infty)$ be a $C^2$ strictly convex function, with $F''$ everywhere positive.
Let $\lambda_n \in [0,1],a_n\le c_0<b_n \in [0,\infty)$ satisfy $$ \lambda_n a_n +(1-\lambda_n)b_n=c_n>c_0 $$ for some constant $c_0$.
Suppose that $c_n \to r>c_0$.
Define $D_n:=F\big(c_n\big)-\lambda_nF(a_n)-(1-\lambda_n)F(b_n) \to 0$, and suppose that $\lim_{n \to \infty}D_n=0$
Must $b_n$ be bounded?
I have a proof for the special case where $c_n=r$ is a constant sequence, but I am having trouble adapting/generalizing it to this case.
Let's make a proof by contradiction.
Claim : Up to taking subsequences, if $a_n$ is bounded and $b_n$ is not, we may assume that $b_n$ tends to infinity, and $a_n$ tends to some point $a$.
Proof : For $a_n$ : Let $I$ be a bounded segment to which $a_n$ belongs for all $n$. Cover $I$ with two segments of half size. By the pigeonhole principle, one of them contains a set $\lbrace a_n | \, n \in D \rbrace$, for $D$ an infinite subset of $\mathbb{N}$. Iterate this construction to this segment. You get a decreasing sequence of segments whose lengths tend to $0$. A fundamental property of the topological space of the real numbers tells you there is a point $a$ in the intersection. Now, for any $\varepsilon > 0$ and $n \in \mathbb{N}$, there exists an index $k > n$ such that $|a_k-a| < \varepsilon$. Indeed, take a small enough segment in the sequence : it contains a set $\lbrace a_n | \, n \in D' \rbrace$, for $D'$ an infinite subset of $\mathbb{N}$, so it contains at least one value $a_k$ for $k$ not in the finite set $\lbrace 0,...,n \rbrace$. Using this property, you can build a subsequence of $(a_n)$ that converges to $a$, by picking increasing indices $n_k$ for which the values $a_{n_k}$ get closer and closer to $a$.
For $b_n$ : Assume you have already extracted a subsequence for $a_n$. You can find $n_{k_1} < n_{k_2} < ... < n_{k_j} < ... $ such that, for all $j$, $b_{n_{k_{j+1}}} > max(( b_{n_i}, \, i \leq j), j)$ The subsequences $(b_{n_{k_{j}}})_j$ and $(a_{n_{k_{j}}})_j$ tend respectively to $\infty$ and $a$.
In fact, the hypotheses of the problem are preserved when we replace the sequences by subsequences. We may also assume (even though it is superfluous at this point) that $\lambda_n$ has a limit. Thus, $\lambda_n$ tends to $1$, since $c_n$ has a finite limit. In fact, we get a better estimate for $b_n$ : $b_n \simeq \frac{r-a}{1-\lambda_n}$.
The hypothesis on $D_n$ tells us the following : $F(r)-F(a)-(1-\lambda_n) F(b_n) \rightarrow 0$. In particular, $F(b_n) = O(\frac{1}{1-\lambda_n}) = O(b_n)$.
By convexity, we show that $F'$ is bounded. By contradiction, let us assume it tends to infinity. Let $M >0$ be an arbitrarily large constant. Let $d > 0 $ such that $F'(x) > M$ whenever $x > d-1$. So, by the mean value theorem, we have $\frac{F(b_n)-F(d)}{b_n - d} > M$ for all sufficiently large $n$. Thus, $F(b_n) > \frac{M}{2} b_n$ for all sufficiently large $n$. This contradicts the previous estimate.
So $F'$ has a limit, let's call it $l$. Let's first deal with the case $l > 0$. I'll write $M_n := \frac{1}{1-\lambda_n}$. So we have : $\frac{F((r-a)M_n(1+o(1)))}{M_n}\rightarrow F(r) - F(a)$. Using the mean value theorem, and the fact $|F(x)| \rightarrow \infty$, we show that $\frac{F((r-a)M_n(1+o(1)))}{M_n}\rightarrow (r-a)l$ :
Declare $\varepsilon >0$. There exists $d$ such that $|F'(x)-l| < \varepsilon $ whenever $x > d-1$. Thus, by the mean value theorem, $|\frac{F((r-a)M_n(1+o(1)))-F(d)}{(r-a)M_n(1+o(1))-d}-l| < \varepsilon$ when $n$ is big enough.
Moreover, since both $F((r-a)M_n(1+o(1)))$ and $(r-a)M_n(1+o(1))$ tend to infinity, if $n$ is big enough, we get $|\frac{F((r-a)M_n(1+o(1)))-F(d)}{(r-a)M_n(1+o(1))-d}-\frac{F((r-a)M_n(1+o(1)))}{(r-a)M_n}| < \varepsilon$. In fact, the quotient of the two fractions tends to $1$, and both fractions are bounded.
Finally, recalling that $(r-a)M_n(1+o(1))$ was our notation for $b_n$, we get $\frac{F(b_n)}{(r-a)M_n} \rightarrow l$.
Thus, $F(r) - F(a) = (r-a) l$. This is not possible, since $F'$ is strictly increasing.
Now, let's assume $l \leq 0$. This implies $F' <0$, so $F$ is strictly decreasing. Thus, $F(r)-F(a) - (1-\lambda_n) F(b_n) < F(r)-F(a) < 0$ for all $n$. So this cannot tend to $0$.
Remark : I don't think you need $F$ to be $C^2$. I believe that the proof works if $F$ is differentiable and strictly convex.