$f(1)-f(a)=\frac{f''(\theta)(1-a)^2}{2}$ Let $f$ be a function on $\mathbb R$ with second derivative and $a\in(0,1)$. If $f(x)\geq f(a)$ and $f''(x)\geq f(a)$ for all $x\in (0,1)$. Then
$$f(1)-f(a)=\frac{f''(\theta)(1-a)^2}{2}$$ and $$f(0)+f(1)\geq \frac{9f(a)}{4}$$.
This question seems to me very obvious in the first look. The first part looks very much like the mean value theorem, but it has a second derivative, where I do not know how to come up with. For the second part, the rough estimate is $f(0)+f(1)\geq 2f(a)$ by passing $f(x)\geq f(a)$ for all $x\in(0,1)$ to limit. But, there is another $\frac{f(a)}{4}$ part, which makes it subtle to me.
Any help will be appreciated.

First note that $f'(a) = 0$ because $f$ has a minimum at $x=a$. Taylor's theorem (with the Lagrange remainder) then gives
$$
 f(x) = f(a) + \frac{(x-a)^2}{2} f''(\theta)
$$
for some $\theta$ between $x$ and $a$, this is the first identity.
Using $f''(\theta) \ge f(a)$ it follows that
$$ 
 f(x) \ge f(a) + \frac{(x-a)^2}{2} f(a)
$$
for all $x \in [0, 1]$, and therefore
$$
 f(0) + f(1) \ge \left( 2 + \frac{a^2}{2} + \frac{(1-a)^2}{2}\right) f(a)
\ge \frac 94 f(a)
$$
since $a^2 + (1-a)^2  = \frac 12 + 2(a - \frac 12)^2 \ge \frac 12$.
