How can I prove $x_{1}+x_{2}>2x_{0}$ Assume that $$f(x)=x^2-2x+\sin{\left(\dfrac{\pi}{2}x\right)},0<x<1$$
The point $x_{0}$ ,such $f(x_{0})=\displaystyle\min_{x\in(0,1)}f(x),$,and the other two  points $x_{1}\neq x_{2}$,such $f(x_{1})=f(x_{2})$
Prove that
$$x_{1}+x_{2}>2x_{0}$$
From the condition, I have $$f'(x_{0})=2x_{0}-2+\dfrac{\pi}{2}\cos{\left(\dfrac{\pi}{2}x_{0}\right)}=0$$
where I first show that 
$$(x_{1}-x_{2})(x_{1}+x_{2}-2)=\sin{\left(\dfrac{\pi}{2}x_{2}\right)}-\sin{\left(\dfrac{\pi}{2}x_{1}\right)}$$
and If I can show
$$\dfrac{\pi}{2}\cos{\left(\dfrac{\pi}{2}x_{0}\right)}>\dfrac{\sin{\left(\dfrac{\pi}{2}x_{1}\right)}-\sin{\left(\dfrac{\pi}{2}x_{2}\right)}}{x_{1}-x_{2}}$$
but I don't see how to make the right estimates.If anyone has an idea how to proceed in that inequality without using Lagrange (the Mean Value Theorem).
 A: first we know there is only one solution $x_0$ for $f'(x)=0$ as in $(0,1) , 2x-2 $is mono increasing and $cos(\dfrac{\pi x}{2})$ is mono decreasing function. WLOG, $x_1<x_2,f(x_1)=f(x_2)\implies x_1<x_0<x_2$
$f'(\dfrac{1}{2})>0 \implies x_0 < \dfrac{1}{2}$
$y_1=x_0-x_1>0,y_2=x_2-x_0>0,x_1+x_2>2x_0 \iff y_2-y_1>0$
edit 1: following equation has a mistake so I need to change the solution:
$f(x_1)=f(x_0-y_1)=f(x_2)=f(x_0+y_2) \iff (y_1+y_2)(y_2-y_1)+2(y_1+y_2)(x_0-1)+2sin(\dfrac{\pi}{2}(\dfrac{y_1+y_2}{4})cos(\dfrac{\pi}{2}(\dfrac{2x_0+y_2-y_1}{4}))=0$
the right one is 
$ y_2^2-y_1^2+2(y_1+y_2)(x_0-1)+2sin(\dfrac{\pi}{2}(\dfrac{y_1+y_2}{2})cos(\dfrac{\pi}{2}(\dfrac{2x_0+y_2-y_1}{2}))=0$
$x_0-1=-\dfrac{\pi}{4} cos\dfrac{\pi x_0}{2}$
let $g(y_1)= y_2^2-y_1^2-(y_1+y_2)\dfrac{\pi}{2} cos\dfrac{\pi x_0}{2}+2sin(\dfrac{\pi}{2}(\dfrac{y_1+y_2}{2})cos(\dfrac{\pi}{2}(\dfrac{2x_0+y_2-y_1}{2}))$
now we prove :$g'(y_1)<0$
$g'(y_1)=-2y_1-\dfrac{\pi}{2} cos\dfrac{\pi x_0}{2}+\dfrac{\pi}{2}cos\dfrac{\pi (y_1-x_0)}{2}=-2y_1+\dfrac{\pi}{2}*2sin\dfrac{\pi y_1}{4}sin\dfrac{\pi (2x_0-y_1)}{4}< -2y_1+\pi *\dfrac{\pi y_1}{4}\dfrac{\pi (2x_0-y_1)}{4}=y_1(\dfrac{\pi ^3(2x_0-y_1)}{16}-2) <y_1(\dfrac{\pi ^3}{16}-2) <0$
$ g(y_1=y_2)=cos\dfrac{\pi x_0}{2}(sin\dfrac{\pi y_2}{2}-\dfrac{\pi y_2}{2})<0 \implies g(y_1)=0 \cap y_1<y_2 $
