Show that if $f(0)=0,f(1)=1$ and $f''\ge0$ then $f(x)\leq x$ for every $x\in[0,1].$ 
Let $f:[0,1]\to\mathbb{R}$ be continuous on $[0,1]$ and twice differentiable on $(0,1)$ such that $\forall x\in(0,1),\ f''(x)\geq0$ .
Suppose also that $f(0)=0,f(1)=1$, show that $\forall x\in[0,1], f(x)\leq x$.

My attempt:
Since the function satisfies the conditions of Lagrange's mean value theorem, then there exists a point $c\in(0,1)$
such that $f'(c)=\frac{f(1)-f(0)}{1-0}=1$.
Now, since the function is twice differentiable then its derivative is also differentiable and therefore continuous on (0,1).

This is where I'm stuck, I'm not sure if that's correct but maybe if I could show that the derivative is continuous on $[0,1]$
I could then use Lagrange's mean value theorem once more and
conclude that there exists a point $c_{2}\in(0,c)$ s.t. $f''(c_{2})=\frac{f'(c)-f'(0)}{c-0}=\frac{1-f'(0)}{c}\overset{f''(x)\geq0}{\overbrace{\geq}}0\Rightarrow f'(0)\leq1.$
Now, using the definition $\lim_{x\to0^{+}}\frac{f(x)-f(0)}{x}\leq1\Rightarrow f(x)\leq x$.
But i'm not sure if that's true because it feels lacking, and even if it's true I don't know how to show that the derivative is continuous on $[0,1]$.
I would appreciate your help, many thanks.
 A: Let $g(x):=f(x)-x,$ so that $g(0)=g(1)=0$ and $g'$ is non-decreasing on $(0,1).$
By contradiction, if $\exists c\in(0,1)\quad g(c)>0$ then, by Lagrange's mean value theorem, $\exists a\in(0,c)\quad g'(a)>0$ and $\exists b\in(c,1)\quad g'(b)<0.$ But this is incompatible with $g'$'s monotonicity.
We thus proved that $\forall x\in[0,1]\quad g(x)\le0,$ i.e. $f(x)\le x.$
A: Since $f''(x)\geq0$ for all $x\in(0,1)$, we know that $f$ is a convex function on $[0,1]$ (see, for example, this question for details). Using convexity we thus have that, for any $x\in[0,1]$,
$$f(x)=f(1x+0(1-x))\leq f(1)x+f(0)(1-x)=x.$$
A: A more direct approach is given by @Lorago. However, if you are familiar with the maximum principles one gets a very quick and slick proof which carries over with little effort to higher dimensions.
Observe that $\Delta(f(x)-x) \geq 0$ on the set $[0,1]$ (here $\Delta$ is just the second derivative in the $x$ variable). The maximum principle now entails that the maximum of $f(x)-x$ lies on the boundary of the domain $[0,1]$. However, $f(0)-0=0=f(1)-1$ and so one concludes that $f(x) - x \leq 0$ on $[0,1]$.
