设f(x)在[0,1]连续,在(0,1)可导,f(0)=0,0<f’(x)<1(x∈(0,1)),求证: [∫01f(x)dx]2>∫01f3(x)dx.

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问题 设f(x)在[0,1]连续,在(0,1)可导,f(0)=0,0<f’(x)<1(x∈(0,1)),求证:
[∫01f(x)dx]2>∫01f3(x)dx.

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答案即证[∫01f(x)dx]2一∫01f3(x)dx>0.考察F(x)=[∫0xf(t)dt]2一∫0xf3(t)dt, 若能证明F(x)>0(x∈(0,1])即可.这可用单调性方法. 令F(x)=[∫0xf(t)dt]2一∫0xf3(t)dt,易知F(x)在[0,1]可导,且 F(0)=0,F’(x)=f(x)[2∫0xf(t)dt—f2(x)]. 由条件知,f(x)在[0,1]单调上升,f(x)>f(0)=0(x∈(0,1]),从而F’(x)与g(x)=2∫0xf(t)dt—f2(x)同号.再考察 g’(x)=2f(x)[1一f’(x)]>0(x∈(0,1)), g(x)在[0,1]连续,于是g(x)在[0,1]单调上升,g(x)>g(0)=0(x∈(0,1]),也就有F’(x)>0(x∈(0,1]),即F(x)在[0,1]单调上升,F(x)>F(0)=0(x∈(0,1]).因此 F(1)=[∫01f(x)dx]2一∫01f3(x)dx>0.即结论成立。

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