$f(c)<1/(b-a)\int^b_a f(x)dxLet f a continous function.
Prove that if f is not constant ,then exists $c,c'\in (a,b)$ such that
$f(c)<1/(b-a)\int^b_a f(x)dx<f(c')$
I tried to apply the mean value theoreme for definite integrals and I get that exists $c\in(a,b)$
$f(c)=1/(b-a)\int^b_a f(x)dx$

As you noted by the mean value theorem there is some $d\in (a,b)$ such that $$f(d)=\frac{1}{b-a}\int_a^bf(x)dx$$
We claim that if $f$ is not constant then $f(d)$ cannot be maximal/minimal, i.e. there are $c,c'\in (a,b)$ such that $f(c)<f(d)<f(c')$, the statement will then follow. Assume that $f(d)$ is minimal, so the function $g:(a,b)\to\Bbb R$, $g(x)=f(x)-f(d)$ is non-negative. We get: $$\int_a^bg(x)dx=\int_a^bf(x)dx-(b-a)f(d)=0$$
As $g$ is non-negative and continuous this implies that $g$ is constant, hence also $f$.
Similarly you can show that $f(d)$ is not maximal if $f$ is non-constant.
