First let's reformulate the question. For $0\leq a\leq c\leq b$, let $\lambda=\lambda(a,c,b)\in[0,1]$ be the number such that $c=\lambda a + (1-\lambda)b$, and for $f\colon \mathbb R_+\to\mathbb R$ define
$$
D_f(a,c,b)= \lambda f(a)+(1-\lambda)f(b)-f(c).
$$
> **Lemma 1.** Let $f\colon \mathbb R_+\to\mathbb R$ be strictly convex and continuously differentiable. Let $0\leq a_n\leq c_n\leq b_n$ be sequences such that $a_n$ is bounded, and $c_n-a_n$ is bounded away from 0. Then, $\lambda(a_n,c_n,b_n)\not \to 0$ implies $D_f(a_n,c_n,b_n)\not\to 0$. 

By taking a sub-sequence on which both $a_n$ converges, Lemma 1 follows from Lemma 2.
> **Lemma 2.** Let $f\colon \mathbb R_+\to\mathbb R$ be strictly convex and continuously differentiable. Let $0\leq a_n\leq c_n\leq b_n$ be sequences such that $a_n\to a<\infty$, and $c_n-a_n$ is bounded away from 0. Then, $\lambda(a_n,c_n,b_n)\not \to 0$ implies $D_f(a_n,c_n,b_n)\not\to 0$.
 
Since it is assumed that $a_n$ and $b_n$ are bounded away, $\lambda(a_n,c_n,b_n)\to 0$ iff $\lambda(a,c_n,b_n)\to 0$ and $\lambda(a_n,c_n,b_n)\to 0$ iff $\lambda(a,c_n,b_n)\to 0$, Lemma 2 follows from Lemma 3.

> **Lemma 3.** Let $f\colon \mathbb R_+\to\mathbb R$ be strictly convex and continuously differentiable. Let $0\leq a\leq c_n\leq b_n$ be sequences such that $c_n-a$ is bounded away from 0. Then, $\lambda(a,c_n,b_n)\not \to 0$ implies $D_f(a_n,c_n,b_n)\not\to 0$.

By applying the affine transformation that maps $a$ to $0$ and $\inf c_n$ to $1$ on the triple $a,c_n,b_n$ and the inverse transformation on $f$, Lemma 3 follows from Lemma 4.

> **Lemma 4.** Let $f\colon \mathbb R_+\to\mathbb R$ be strictly convex and continuously differentiable. Let $1\leq \leq c_n\leq b_n$ be sequences. Then, $\lambda(0,c_n,b_n)\not \to 0$ implies $D_f(0,c_n,b_n)\not\to 0$.

By taking a sub-sequnce on which $\lambda(0,c_n,b_n)$ is bounded away from 0, Lemma 4 follows from Lemma 5.

> **Lemma 5.** Let $F\colon \mathbb R_+\to\mathbb R$ be strictly convex and continuously differentiable. Let $1\leq \leq c_n\leq b_n$ be sequences. Then, $inf\lambda(0,c_n,b_n)>0 $ implies $\inf D_F(0,c_n,b_n)> 0$.

**Proof of Lemma 5.**
We assume wlog that $F(0)=0$ and denote $\lambda_n=\lambda(0,c_n,b_n)$ and $D_n=D_F(0,c_n,b_n)$.

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 6.** $G$ is positive and increasing in both $x$ and $y$.

**Proof of Claim 6.**
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 Claim 6.

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