approximation by means of simplify fraction

Let f(x) is continuous function on [a,b].
\rho_n (x) = \sum\limits_{k=1}^n 1/{x-t_k} - the simplest fraction of best approximation for f(x)
i.e. E_n(f)=max|f(x)-\rho_n (x)| (on [a,b]) where E_n - the smallest deviation for f(x).
t_k - some real numbers.
THEN exists more than n point of Chebyshev alternation (i.e. >=(n+1) points) i.e. exists points:
a <= x_1 < x_2 < ... < x_{n+1} <=b : \rho_n (x_j) - f(x_j) = (-1)^{j+1}*E_n(f) for j=1,...,n+1.

Help me to prove this theorem or offer idea's.

Thanks.

P.S: Chebyshev proved that exist polynomial P_n(x) of best approximation for f(x) that exist >=(n+2)
point of Chebyshev alternation!
.