Estimate the arc length of the graph of a particular $\mathcal{C}^1$ function from $[0,1]\to [0,1]$. Let $f:[0,1]\to[0,1]$ be $\mathcal{C}^1$ such that $f(0) = f(1) = 0$ and $f'$ is nonincreasing ($f$ concave). Show that the arc length of the graph is smaller than 3.
I have a rather geometric proof. For any $P$ be any (finite) partition of $[0,1]$, then
$$\Lambda(P,(x,f(x)))$$
is nothing but the sum of length of finitely many polygon arcs. Using mean value theorem we can show that these arcs are convex.

By certain argument we can show that sum of the lengths of these polygon arcs are always smaller than sum of the boundary three sides, which is 3.
Since this is true for any $P$ and $$\Gamma((x,f(x)) = \sup_P\Gamma(P,(x,f(x)))\le 3.$$
We are done.
I want to know a purely analytic proof, which just estimates
$$\Gamma((x,f(x)) = \int_0^1 \sqrt{1+f'(x)^2}dx$$ using the information given.
 A: 
Lemma. Let $f:[0,1]\to\mathbb{R}$ be a concave and continuously differentiable function, with $f(0)=f(1)=0$. Then
  $$\int_0^1\sqrt{1+f'^2(x)}dx\leq 1+2\Vert f\Vert_\infty.$$
  where $\Vert f\Vert_\infty=\sup_{x\in[0,1]}|f(x)|.$

Proof. Using the simple inequality $\sqrt{1+t^2}\leq 1+|t|$ we conclude that
$$
\int_0^1 \sqrt{1+f'^2(x)}\,dx\leq1+\int_0^1|f'(x)|\,dx
$$
Now, $f'$ is non-increasing, so there is $x_0\in[0,1]$ such that $f' \geq0$ on $[0,x_0]$ and $f' \leq0$ on $[x_0,1]$, ($x_0$ might be $0$ or $1$), thus
$$
\int_0^1|f'(x)|\,dx=\int_0^{x_0} f'(x) \,dx-\int_{x_0}^1 f'(x) \,dx
\leq f(x_0)+f(x_0)=2f(x_0)\leq 2\Vert f\Vert_\infty
$$
and the lemma follows.
Now, if $f([0,1])\subset[0,1]$, we have $\Vert f\Vert_\infty\leq 1$ and the desired inequality follows.$\qquad\square$
Remark. With the same proof, if $f:[0,1]\to\mathbb{R}$ is a concave and continuously differentiable function. Then
$$\int_0^1\sqrt{1+f'^2(x)}dx\leq 2\Vert f\Vert_\infty-f(0)-f(1).$$
A: We construct a sequence $(\Gamma_n)_{n\geq0}$ of polygonal arcs approximating the graph $\gamma$ of $f$ as follows:
$\Gamma_0$ connects the points $(0,0)$, $(0,1)$, $(1,1)$, and $(1,0)$ in turn. Drawing a horizontal upper supporting line of $\gamma$, cutting off a rectangle from $\Gamma_0$, we obtain $\Gamma_1$. Drawing two supporting lines having inclinations $\pm45^\circ$ with respect to the horizontal we then cut off two triangles from $\Gamma_1$ and obtain $\Gamma_2$. Drawing four supporting lines having inclinations $\pm67.5^\circ$, $\>\pm22.5^\circ$ with respect to the horizontal we then cut off four triangles from $\Gamma_2$ and obtain $\Gamma_3$;  see the following figure:

In general $\Gamma_n$ is obtained from $\Gamma_{n-1}$ by drawing $2^{n-1}$ supporting lines having as inclinations  the odd multiples of ${\pi\over 2^n}$ in the interval $\bigl[-{\pi\over2},{\pi\over2}\bigr]$. These lines are cutting off $2^{n-1}$ triangles from $\Gamma_{n-1}$. It is obvious that ${\rm length}(\Gamma_n)\leq{\rm length}(\Gamma_{n-1})$, and as $\gamma$ is supposed to be smooth one has $$\lim_{n\to\infty}{\rm length}(\Gamma_n)={\rm length}(\gamma)\ .$$
It follows that $${\rm length}(\gamma)\leq{\rm length}(\Gamma_0)=3\ .$$
