Bound on Derivative for Twice Differentiable Convex Function

We are given that $f: \mathbb{R} \to \mathbb{R}$ is a twice differentiable function such that $f''(x) > 0$ for all $x \in \mathbb{R}$.

1. Monotonicity of the first derivative: Since $f''(x) > 0$ for all $x \in \mathbb{R}$, the first derivative $f'(x)$ is a strictly increasing function on $\mathbb{R}$.

2. Application of Mean Value Theorem: The function $f$ is continuous on the closed interval $\left[\frac{1}{2}, 1\right]$ and differentiable on the open interval $\left(\frac{1}{2}, 1\right)$. By Lagrange's Mean Value Theorem, there exists at least one point $c \in \left(\frac{1}{2}, 1\right)$ such that: $f'(c) = \frac{f(1) - f\left(\frac{1}{2}\right)}{1 - \frac{1}{2}}$

3. Evaluation of $f'(c)$: Substituting the given values $f(1) = 1$ and $f\left(\frac{1}{2}\right) = \frac{1}{2}$: $f'(c) = \frac{1 - \frac{1}{2}}{1 - \frac{1}{2}} = 1$

4. Bounding $f'(1)$: Since $c \in \left(\frac{1}{2}, 1\right)$, we have $c < 1$. Because $f'(x)$ is strictly increasing, $c < 1$ implies: $f'(c) < f'(1)$

   Substituting $f'(c) = 1$ into the inequality yields: $1 < f'(1) \implies f'(1) > 1$

Thus, the correct inequality is $f'(1) > 1$, which corresponds to Option D.