Continuous function of one variable Let $f(x)$ continuous function on $R$ wich can be in different signs. Prove, that there is exists an arithmetic progression $a, b, c   (a<b<c)$,  such that $f(a)+f(b)+f(c)=0$.

Let's ponder like this:
At some point $x$ $f(x)>0$, therefore, in the vicinity of this point there is an increasing arithmetic progression
$a_{0}, \ b_{0}, \ c_{0}$ that, $f(a_{0})+f(b_{0})+f(c_{0})>0$. 
Like this one will be found increasing arithmetic progression of 
$a_{1}, \ b_{1}, \ c_{1}$ that, $f(a_{1})+f(b_{1})+f(c_{1})<0$. 
For all values of parametr $t[0,1]$ сonsider the arithmetic progression 
$a(t), \ b(t), \ c(t)$, where 
$a(t)=a_{0}(1-t)+a_{1}t$,  
$b(t)=b_{0}(1-t)+b_{1}t$, 
$c(t)=c_{0}(1-t)+c_{1}t$. 
Function $F(t)=f(a(t))+f(b(t))+f(c(t))$ continuously depends on $t$, at $t=0 \ F(t)>0$, and at $t=1 \ F(t)<0$. It means that in some $t \ F(t)=0$ and the corresponding progress  $a(t), \ b(t), \ c(t)$  is required. 
