An inequality related to convex function Let $f$ be twice differentiable on $[a,b]$, and $f''(x)\geq 0$, $f(a)\leq 0$, $f(b)\leq 0$. Prove then $$f(x)\geq \frac{2}{b-a}\int_a^b f(t)\, dt,\quad \forall\ x\in [a,b].$$
I do not know how to use $f(t)\leq 0$, $\forall t$.

Let $g=-f$. Then $g(a),g(b)\ge 0$, $g^{\prime\prime}(x)\le 0$ and we want prove $g(x)\le \dfrac{2}{b-a}\int_{a}^bg(t)dt$.
Fix $x\in(a,b). $Let $G_x:[a,b]\to\mathbb{R}$ given by $G_x(t)=\left\{\begin{array}{ccc}\frac{g(x)}{x-a}(t-a)&\mbox{if}&0\le t<x\\\frac{g(x)}{x-b}(t-b)&\mbox{if}&x\le t\le b\end{array}\right.$
Clearly $G$ is continuos. More over, 
$\begin{eqnarray}\int_a^bG_x(t)dt&=&\int_0^x\frac{g(x)}{x-a}(t-a)dt+\int_x^b\frac{g(x)}{x-b}(t-b)dt\\
&=&\left.\frac{g(x)}{x-a}\frac{(t-a)^2}{2}\right|_{t=a}^x+\left.\frac{g(x)}{x-b}\frac{(t-b)^2}{2}\right|_{t=x}^b\\
&=&\dfrac{g(x)(x-a)}{2}-\frac{g(x)(x-b)}{2}\\
&=&\dfrac{(b-a)g(x)}{2}
\end{eqnarray}$
Now, since $g^{\prime\prime}\le 0$, then $g$ is concave. Thus, for each $x\in (a,b)$, the segment joining points $(a,g(a))$ and $(x,g(x))$ is under the graph of $g$ from $a$ to $x$. But this segment is the graph of $G_x$ for $t\in[a,x]$. Thus, $g(t)\ge G_x(t)$ for al $t\in[a,x]$.
By the same way, $g(t)\ge G_x(t)$ for al $t\in[x,b]$. In deffinitive $G_x(t)\le g(t)$  for all $t\in[a,b]$. Integrating both sides respect to $t$ give us the desired inequality for all $x\in(a,b)$.
For $x=a$ or $x=b$, the inequility follows by continuity.
