$f$ is differentiable on $[0,+\infty)$, $f'(x)$ is strictly non-decreasing on $[0,+\infty)$ $f$ is differentiable on $[0,+\infty)$, $f'(x)$ is strictly non-decreasing on $[0,+\infty)$. Prove
$$\int_0^{2x} f(t) \text dt \ge 2xf(x) , \forall x\in[0,+\infty)$$
My intuition is construct a function like $F(x)=\int_0^{2x} f(t) \text dt-2xf(x)$. Then $F'(x)=2f(2x)-2f(x)-2xf'(x)$, I want to discuss the behavior of $F'(x)$ and see if I can get some information to make  $F(x)\ge 0$. But it seems not possible. Maybe I should consider Taylor series or Mean Value Theorem. But get no clues...
 A: Note that
$$\int_0^{2x} f(t) \, dt - 2xf(x) = \int_x^{2x} [f(t) - f(x)] \, dt - \int_0^x [f(x) - f(t)] \, dt \\ = \int_0^{x} [f(x+t) - f(x)] \, dt - \int_0^x [f(x) - f(x-t)] \, dt, $$
where the variable changes $t \to x+t$ and $t \to x-t$ have been applied for the first and second integrals, respectively, on the RHS.
Applying the MVT, we have $x < \theta_t < x+t$ and $x-t < \eta_t < x$ such that
$$f(x+t) - f(x) = tf'(\theta_t) \geqslant tf'(\eta_t) = f(x) - f(x-t),$$
for all $t \in [0,x]$ since $f'(\theta_t) \geqslant f'(\eta_t).$
Thus, 
$$\int_0^{2x} f(t) \, dt - 2xf(x)  = \int_0^x \{[f(x+t) - f(x)] - [f(x) - f(x-t)] \}  \, dt \geqslant 0,$$
A: When $x>0$,
$$F'(x) =2x\left(\frac{f(2x)-f(x)}{x} -f'(x)\right)$$
Applying mean value theorem on the interval $[x,2x]$, $\exists\, c\in (x, 2x)$ such  that $$f'(c) =\frac{f(2x)-f(x)}{x}$$
Now since $f'(x)$ is an increasing sequence  and $c>x$, we can say $f'(c) >f'(x)$. 
$F'(x)$ is an increasing  sequence  so $F(x)>F(0)\quad \forall\, x>0$.
A: The condition translates to $f$ being (strictly) convex. 
The integral can be rearranged in the following way: 
$$\int\limits_{0}^{2x} (f(t)-f(x)) \, dt \geq 0$$
Draw a picture of a convex function, and you will see immediately why this is true, even with strict inequality. (Hint: "pair up" the values $t$ and $2x-t$).
A: Since $f$ is a (strictly) convex function, its graph lies above the tangent line at $(x, f(x))$.
Hence the integral $\int_0^{2x} f$ is greater than the area of a trapezoid whose area is exactly  $2x f(x)$.
(Here $2x$ is the length of the base and $f(x)$ is the height at the midpoint.)
