How can I prove this inequality? $c\leq x\leq a+b$ 
Show that if ABC is a right triangle and curve $x$ is inside the ABC triangle then we have
$$c\leq x\leq a+b$$
Do not use integral.
 A: The inequality $c\leq x$ is just the statement "the shortest distance between any two points is a straight line." The upper bound $x\leq a+b$ is harder.
So far as I know, there isn't an elementary geometric argument. Suppose the graph curve is rectifiable (let's say with arclength $s$ rather $x$), and let $\gamma(x) = (x,f(x))$ parameterize the curve. Given $\epsilon>0$, the definition of rectifiability implies there are points $0=x_0<x_1<\dotsm< x_n=b$ such that
$$s-\epsilon<\sum_{i=0}^{n-1}d(\gamma(x_i),\gamma(x_{i+1}))$$
where $d$ is distance in $\mathbb{R}^2$. By triangle inequality we have
$$d(\gamma(x_i),\gamma(x_{i+1})) = \sqrt{(x_i-x_{i+1})^2+(f(x_i)-f(x_{i+1})^2}\leq |x_i-x_{i+1}|+|f(x_i)-f(x_{i+1})|.$$
Moreover, a continuous injective function is monotone (and hence decreasing in our case), so $|x_i-x_{i+1}|=x_{i+1}-x_i$ and $|f(x_i)-f(x_{i+1})| = f(x_i)-f(x_{i+1})$. Thus
$$s-\epsilon < \sum_{i=0}^{n-1}d(\gamma(x_i),\gamma(x_{i+1}))\leq\sum_{i=0}^{n-1}(x_{i+1}-x_i)+\sum_{i=0}^{n-1}\big(f(x_i)-f(x_{i+1})\big) = b+a.$$
Therefore $s<a+b+\epsilon$. Since $\epsilon>0$ was arbitrary in our argument, we conclude $s\leq a+b$, as desired.
As an aside, I believe the upper bound should be achievable with a construction similar to that of the Cantor function (whose graph in the unit square is nondecreasing and has length 2).
