Vtyjkyuk

$f$ is differentiable on $[0,+infty)$, $f'(x)$ is strictly non-decreasing on $[0,+infty)$. Prove
$$int_0^2x f(t) text dt ge 2xf(x) , forall xin[0,+infty)$$

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...

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,

$$int_0^2x f(t) , dt - 2xf(x) = int_0^x [f(x+t) - f(x)] - [f(x) - f(x-t)] , dt geqslant 0,$$

When $x>0$,
$$F'(x) =2xleft(fracf(2x)-f(x)x -f'(x)right)$$
Applying mean value theorem on the interval $[x,2x]$, $exists, cin (x, 2x)$ such that $$f'(c) =fracf(2x)-f(x)x$$

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 $F(x)>F(0)quad forall, x>0$.

The condition translates to $f$ being (strictly) convex.
The integral can be rearranged in the following way:
$$intlimits_0^2x (f(t)-f(x)) , dt geq 0$$

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 $2x-t$).

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.)