設(shè)g(x) = [f(x)]^3[f(1-x)]^4
則,g(x)在[0,1]連續(xù),在(0,1)可導(dǎo).
g(0) = [f(0)]^3[f(1)]^4 = 0,
g(1) = [f(1)]^3[f(0)]^4 = 0 = g(0).
g'(x) = 3[f(x)]^2f'(x)[f(1-x)]^4 + 4[f(x)]^3[f(1-x)]^3f'(1-x)(-1)
= [f(x)]^2[f(1-x)]^3{3f'(x)f(1-x) - 4f(x)f'(1-x)}
由羅爾中值定理,至少存在一點(diǎn)ξ∈(0,1)使得
g'(ξ)=[f(ξ)]^2[f(1-ξ)]^3{3f'(ξ)f(1-ξ) - 4f(ξ)f'(1-ξ)} = 0,
但由于,在（0,1）內(nèi)f（x）!=0.
因此,[f(ξ)]^2[f(1-ξ)]^3 不等于0,
故,必有,
3f'(ξ)f(1-ξ) - 4f(ξ)f'(1-ξ) = 0,
成立.
也就是,
3f'(ξ)f(1-ξ) = 4f(ξ)f'(1-ξ),
因此,有
3f'(ξ)/f(ξ) = 4f'(1-ξ)/f(1-ξ).
所以命題得證.