Will f(x) in x =x? According to Taylor formula of Lagrange remainder, it is expanded to n= 1.

f(x)=f(x？ )+f'(x？ )*(x- x？ )+f''(ξ)/2！ *(x- x？ )?

Note that f'(a)=f'(b)=0.

Take x=(a+b)/2, take x? =a and x? =b substitution, get

f[(a+b)/2]= f(a)+f '(a)*[(b-a)/2]+f ' '(ξ？ )/2! *[(b-a)/2)？ (a & ltξ? & lt(a+b)/2) ①

f[(a+b)/2]= f(b)+f '(b)*[(a-b)/2]+f ' '(ξ？ )/2! *[(a-b)/2)？ ((a+b)/2 & lt； ξ? & ltb) ②

②-① Finishing

1/2*[f''(ξ？ )-f''(ξ？ )]= -4*[f(b)-f(a)]/ (b-a)？

And max{|f''(ξ? )|，|f''(ξ？ )|}≥ 1/2*[|f''(ξ？ )|+|f''(ξ？ )|]≥ 1/2*|f''(ξ？ )-f''(ξ？ )|=4*|f(b)-f(a)|/ (b-a)？

Because of the existence of c∈(a, b), |f''(c)|= max{|f''(ξ? )|，|f''(ξ？ )|}

So |f''(c)|≥4*|f(b)-f(a)|/ (b-a)?