Length of a curve defined by a convex function Let $f \colon [0,1] \to [0,1]$ be a function of class $C^1$ such that $f(0)=f(1)=1$ and $f'$ is nondecreasing, i.e., $f$ is convex. Show that the length of the curve defined by the graph of $f$ is less than or equal to 3.
 A: The intuition behind the result becomes more apparent if the graph of $f$ is adjoined with the segment between points $(0,1)$ and $(1,1)$ as to form a closed curve $\,\Gamma\,$. It is quite obvious that $\,\Gamma\,$ is a convex curve, and that it is enclosed in the unit square with opposite corners at $(0,0)$ and $(1,1)$. Therefore, the perimeter of $\,\Gamma\,$ can be no larger than the perimeter of the enclosing square, which is equivalent to the problem statement.
This intuition can be formalized as shown in these answers on MO, with the latter quoting Archimedes for what must be the earliest formulation of the principle:

If two plane  curves  C  and  D  with the same  endpoints  are  concave  in the same  direction,  and  C  is  included  between  D  and  the  straight line joining  the  endpoints,  then  the  length  of  C  is less  than  the  length  D. 

A: Notation : $[xy]$ is a line segment between $x$ and $y$
$[xyz]$ is a triangle whose vertexes are $x,\ y$, and $z$. 
Proof : Assume that $f(x_0)\leq f(x)$ for all $x$
If $t_i$ is a partition s.t.
$$ A_i=(t_i,f(t_i)),\ \sum_{i=0}^{n-1}\ |A_i-A_{i+1}| +\varepsilon > {\rm
length \ of \ graph\ of}\ f$$
and $t_i=x_0$ for some $i$, then $$\sum_{j=1}^i\ \| A_{j-1}
 -A_j\|_1 \geq \sum_{j=1}^i\ |A_{j-1} -A_j| $$
where $\|\ \|_1$ is Manhattan distance and $$\sum_{j=i}^n\ \| A_{j}
 -A_{j+1}\|_1 \geq \sum_{j=i}^n\ |A_{j} - A_{j+1} | $$ Here $ \sum_{j}\ \| A_{j-1}
 -A_j\|_1 \leq 3$ so that the prooof is followed.
Proof using Euclidean distance : For convenience, we assume that $f(\frac{1}{2}=x_0)=0$.
Then consider $$ S:=\bigg\{ (0,t)\bigg|0\leq t\leq 1\bigg\} \cup \bigg\{ (t,0) \bigg|0\leq
t\leq \frac{1}{2} \bigg\} ,\ T:=\sum_{j=1}^i\ [A_{j-1} A_j] $$$$
Consider a region $R$ enclosed by $S\cup T$. We cut $R$ into $R_2$
and a piece $Q_2$. We cut $R_2$ into $R_3$ and $Q_3$. If we repeat
the process, so $R_i$ goes to $T$ in Hausdorff distance. 
Here we can assume that all $Q_i=[a_ib_ic_i]$ are triangle with
$[b_ic_i]\subset R_i$. Since $$ |a_i-b_i|+|a_i-c_i|\geq |b_i-c_i|$$
so length of $\partial R_i-T$ is decreasing.
