How to prove this inequality about the arc-length of convex functions?

Let be $f,F:[0,1] \to \mathbb{R}$ with $f,F \in C^2([0,1])$ two convex functions such that $f \le F$ in all points and $f(0)=F(0)$, $f(1)=F(1)$. Considering the plane-curves $\gamma(t)=(t,f(t))$ and $\Gamma(t)=(t,F(t)), t\in[0,1]$ prove that the arc length of $\gamma$ is equal or greater than $\Gamma$'s one.

Using length formula for a cartesian curve, this question became the following: demonstrate \begin{equation} \int_0^1 \sqrt{1+\{f'(t)\}^{2}}dt \ge \int_0^1 \sqrt{1+\{F'(t)\}^{2}}dt \end{equation}

Thanks for any help.

Take $\psi(t)=(1+t^2)^{\frac 1 2}$, then $\psi''(t)=(1+t^2)^{-\frac 3 2}\geq 0$, $\psi$ is convex on $\mathbb{R}$. By Taylor-lagrange : $$\psi(t)=\psi(s)+(t-s)\psi'(s) +\frac 1 2 (t-s)^2 \psi''(\xi) \geq \psi(s)+(t-s)\psi'(s)~~ \forall t,s \in \mathbb{R}$$

So, for all $t \in [0,1]$ :

$$(1+f'(t)^2)^{\frac 1 2} \geq (1+F'(t)^2)^{\frac 1 2} + (f'(t)-F'(t)) \psi'(F'(t))$$

We integrate that inequality :

$$\int_0^1(1+f'(t)^2)^{\frac 1 2}dt\geq \int_0^1(1+F'(t)^2)^{\frac 1 2} dt+ \int_0^1(f'(t)-F'(t)) \psi'(F'(t))dt$$

Let us prove that $\int_0^1(f'(t)-F'(t)) \psi'(F'(t))dt \geq 0$.

Since $f,F \in C^2([0,1])$, we can integrate by part, then :

$$\int_0^1(f'(t)-F'(t)) \psi'(F'(t))dt=\left[ (f(t)-F(t)) \psi'(F'(t))\right]^1_0 -\int_0^1(f(t)-F(t)) \psi''(F'(t))F''(t)dt$$

So : $$\int_0^1(f'(t)-F'(t)) \psi'(F'(t))dt=0- \int_0^1(f(t)-F(t)) \psi''(F'(t))F''(t)dt \geq 0$$

Because $f \leq F$, $\psi'' \geq 0$ and $F'' \geq 0$