Prove that $f''(x) \ge 4$ for some $x \in \left[0,\frac12\right]$. 
Suppose $f$ is a twice-differentiable function with $f(0) = 0$, $f\left(\frac12\right) = \frac12$ and $f'(0) = 0$. Prove that $f''(x) \ge 4$ for some $x \in \left[0,\frac12\right]$.

Below is my proof, and I was wondering if what I did in the bolded part especially is right and if my proof is enough (i.e. if I have to find a case where $f''(x)>4$ not just $f''(x)=4$.

The Mean Value Theorem (MVT) states that if there is some function $f(x)$ that is continuous over the interval $[a, b]$ and differentiable on $(a,b)$, then there exists a real number $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$. Since $f$ in this problem is twice-differentiable, $f$, $f'$ and $f''$ are all continuous and differentiable over any interval where they are defined. Specifically, $f$ and its derivatives are defined over the interval $(0,\frac{1}{2})$ since differentiability implies continuity. 
Now let $g(x) = f(x) - 2x^2$. Then $$g(0) = f(0) - 2(0)^2 = 0$$ and $$g(\frac{1}{2}) = f(\frac{1}{2}) - 2(\frac{1}{2})^2 = \frac{1}{2} - \frac{1}{2} = 0,$$ using the values $f(0) = 0$ and $f\left(\frac12\right) = \frac12$  given.
  Next, $g'(x) = f'(x) - 4x$. Therefore, $g'(0) = f'(0) - 0$.
  Also, by MVT, there exists some $n$ where $0<n<\frac{1}{2}$ such that $$g'(n) = \frac{g(\frac{1}{2}) - g(0)}{\frac{1}{2} - 0} = 0.$$ 
  So using the MVT again, there exists some number $m$ where $0<m<n$ such that $$g''(m) = \frac{g'(n) - g'(0)}{n} = 0.$$ 
Since we found $g''(m) = 0$, let us differentiate $g'(x)$: $g''(x) = f''(x) - 4$. Therefore, $$f''(x) = g''(x) + 4 .$$ Applying this to the value $m$, $f''(m) = g''(m) + 4$ — however, we already found that $g''(m) = 0$, so $f''(m) = 4$. And since $0<m<n$ while $0<n<\frac{1}{2}$, $0<m<\frac{1}{2}$. Therefore, we have found a value for some $x \in \left[0,\frac12\right]$ that $f''(x) \ge 4.$ No further proof is needed since equality is part of the condition and we have already found the equality case.

 A: 
I was wondering if what I did in the bolded part especially is right and if my proof is enough (i.e. if I have to find a case where $f''(x)>4$ not just $f''(x)=4$.

You cannot prove what is wrong. Look at $f(x) = 2x^2$. It satisfies the condition, but its twice-derivative is $4$ everywhere.
Finding a value $x$ where $f''(x) = 4$ is enough for the problem.

However, there is some problem with this:

The Mean Value Theorem (MVT) states that if there is some function $f(x)$ that is continuous over the interval $[a, b]$ and differentiable on $(a,b)$, then there exists a real number $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$. Since $f$ in this problem is twice-differentiable, $f$, $f'$ and $f''$ are all continuous and differentiable over any interval where they are defined. Specifically, $f$ and its derivatives are defined over the interval $(0,\frac{1}{2})$ since differentiability implies continuity. 

No, $f''$ does not need to be continuous or differentiable. However, this does not affect the other parts of your proof.
