Prove that there are $0
Given $g$ is continuous on $[0,1]$ and differentiable on  $(0,1)$,
  $g(0)=0$, and $g(1)=1$.  
1) Prove that there exist $0<c_1<c_2<1$ such that
  $$\frac{1}{g'(c_1)}+\frac{1}{g'(c_2)}=2.$$ 
2) Prove that there exist $0<c_1<c_2<1$ such that 
  $$g'(c_1)g'(c_2)=1.$$
I am thinking if Mean Value Theorem would help, but Mean Value theorem could only prove that in interval (0,1) there exists $g'(c)=1$.
 A: For the first part see Prove there exist $0\le c_1<c_2<\dots<c_n\le1$ where $\sum_{k=1}^n\frac1{f'(c_k)}=n$
Now we show that there are $0< c_1<c_2< 1$ such that $g'(c_1) g'(c_2)=1$. 
Since $g(1)-g(0)=1$, by the MVT there is at least one $c\in (0,1)$ such that $g'(c)=1$. We may assume that $c$ is the only one, otherwise the claim is trivially true. We have three cases.
1) $g'(x)\leq 1$ on $(0,1)$. Then we should have $g'(x)<1$ in $(0,c)$ and $(c,1)$ and by the MVT, 
$$g(c)-g(0)<c-0\;,\;g(1)-g(c)<1-c\implies 1=g(1)-g(0)<(1-c)+c=1$$
Contradiction.
2) $g'(x)\geq 1$ on $(0,1)$. Then we should have $g'(x)>1$ in $(0,c)$ and $(c,1)$ and by the MVT, 
$$g(c)-g(0)>c-0\;,\;g(1)-g(c)>1-c\implies 1=g(1)-g(0)>(1-c)+c=1$$
Contradiction.
3) There are $s,t\in (0,1)$ such that $a=g'(t)<1$ and $b=g'(s)>1$.
Then, by Darboux theorem, $g'$ assumes between $s$ and $t$ all the values in $(1-r,1+r)\subset (g'(t),g'(s))$ for some $r\in (0,1)$.
Now note that $1-r<\frac{1}{1+\frac{r}{2}}<1+\frac{r}{2}<1+r$, hence there exist $c_1,c_2\in (0,1)$ such that 
$$g'(c_1)g'(c_2)=\frac{1}{1+\frac{r}{2}}\cdot\left(1+\frac{r}{2}\right)=1$$
and we are done.
