1. Given conditions

We are given that for all $x\in(a,b)$:

• $f'(x)>0$ $\Rightarrow$ $f$ is increasing,
• $f''(x)<0$ $\Rightarrow$ $f$ is concave downward.

We must find a quantity that is always less than $\frac{f(c)-f(a)}{f(b)-f(c)}$ for any $c\in(a,b)$.

2. Use Mean Value Theorem on the two intervals

Apply MVT on $[a,c]$. Since $f$ is continuous on $[a,c]$ and differentiable on $(a,c)$, there exists $\xi_1\in(a,c)$ such that

So,

Similarly, apply MVT on $[c,b]$. There exists $\xi_2\in(c,b)$ such that

so

Hence,

3. Use concavity: $f''(x)<0$

Since $f''(x)<0$, the derivative $f'(x)$ is strictly decreasing on $(a,b)$.

Also, because $a<\xi_1<c<\xi_2<b,$ we get $f'(\xi_1)>f'(\xi_2).$

Therefore,

So,

Thus,

4. Check options

We have proved that the expression is always greater than $\frac{c-a}{b-c}.$ This is exactly Option D.

Let us briefly inspect others:

• A: $1$ — not always true as the ratio can be less than $1$ depending on $c$.
• B: $\frac{b-c}{c-a}$ — not guaranteed.
• C: $\frac{b+a}{b-a}$ — unrelated to the concavity argument.
• D: $\frac{c-a}{b-c}$ — always true.

5. Final answer

So the correct option is D.