<s>[INST] Prove that there exists a point $c$ such that $f''(c)\ge 4$ 

Let $f:[0,1/2]\to\mathbb{R}$ be a twice differentiable function such that $f(0) = 0$ and $f'(0) = 0$. Also $f(1/2)\ge 1/2$. Prove that there exists a $c\in[0,1/2]$ such that $f''(c)\ge4$.


I have tried different approached. By Lagrange's MVT twice we can prove the that there is a $c$ such that $f''(c)\ge 2$.
One of my second approaches involves taking two cases $c\le1/4$ and $c>1/4$. Another approach was to consider $g(x) = f(x) + f(x+1/4)$ or $g(x) = f(x)-f(x-1/4)$. 
 [/INST] We have 3 data points, $f(0) = f(0+\delta) = 0$ and $f(\tfrac{1}{2}) \geq \tfrac{1}{2}$.  Using 3 points, we can try to estimate the 2nd derivative.

For any 3 data points, $a \mapsto f(a), b \mapsto f(b), c \mapsto f(c)$, the interpolating quadratic polynomial should be: 
$$ f(x) \approx f(a)\frac{(x-b)(b-c)(c-x)}{(a-b)(b-c)(c-a)}
+ f(b)\frac{(a-x)(x-c)(c-a)}{(a-b)(b-c)(c-a)}
+ f(c)\frac{(a-b)(b-x)(x-a)}{(a-b)(b-c)(c-a)}
$$
Notice the cancellation, so we can just cross out the factors in numerator and denominator:
$$ f(x) \approx \frac{(x-b)(c-x)}{(a-b)(c-a)}
+ \frac{(a-x)(x-c)}{(a-b)(b-c)}
+ \frac{(b-x)(x-a)}{(b-c)(c-a)}
$$
This is just for any 3 points, there is a parabola going through it.
https://stackoverflow.com/questions/717762/how-to-calculate-the-vertex-of-a-parabola-given-three-points

In our case: $a=0,b = 0 + \delta, c = \tfrac{1}{2}$:
$$ f(x) \approx 
  0\cdot \frac{(x-\delta)(\tfrac{1}{2}-x)}{(-\delta )\tfrac{1}{2}}
+ 0\cdot \frac{(-x)(x-\tfrac{1}{2})}{(-\delta)(\delta-\tfrac{1}{2})}
+ (\tfrac{1}{2} + \epsilon) \cdot \frac{(\delta-x)x}{(\delta-\tfrac{1}{2})(\tfrac{1}{2})}
\approx \boxed{(2+\epsilon)x^2 }
$$
with $\epsilon > 0$.  In this case the 2nd derivative should be something like:  $f''(x) \approx 4 + \epsilon$.
The Mean Value Theorem should tell us this point with $f''(c) = 4$ is a topological feature and can't be removed.

See section 4 of Osculating curves: around the Tait-Kneser Theorem by Ghys, Tabachnikov and Timorin.
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