设f(x)在[0,1]上连续,在(0,1)内可导,f(0)=0,0≤f’(x)≤1,求证:[∫01f(x)dx]2≥∫01</subp>f3(x)dx.

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问题 设f(x)在[0,1]上连续,在(0,1)内可导,f(0)=0,0≤f’(x)≤1,求证:[∫01f(x)dx]2≥∫01</subp>f3(x)dx.

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答案首先证明不等式 [∫0xf(t)dt]2≥∫0xf3(t)dt(0≤x≤1). 令F(x)=[∫0xf(t)dt]-∫0xf3(t)dt, F’(x)=2∫0xxf(t)dt.f(x)-f3(x) =f(x)[2∫0xf(t)dt-f2(x)], 再令φ(x)=2∫0xf(t)dt-f2(x)则 φ’(x)=2f(x)-2f(x)f’(x) =2f(x)[1-f’(x)]. 因为f(0)=0,f’(x)≥0,所以f(x)单增,当x≥0时,f(x)≥f(0)=0. 又0≤f’(x)≤1,于是φ’(x)≥0,由此φ(x)单增,当x≥0时,φ(x)≥φ(0)=0, 所以又有F’(x)≥0,由此F(x)单增,当x≥0时,F(x)≥F(0)=0,故F(1)≥0, 从而有[∫01f(x)dx]2≥∫01(x)dx.

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