设函数f(x)具有二阶导数,且f”(x)>0,又函数u(x)在区间[0,a]上连续,证明: ∫0af[u(t)]dt≥f[∫0au(t)dt].

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问题 设函数f(x)具有二阶导数,且f”(x)>0,又函数u(x)在区间[0,a]上连续,证明:
    0af[u(t)]dt≥f[0au(t)dt].

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答案注意到f”(x)>0,由泰勒公式,得 f(x)=f(x0)+f’(x0)(x-x0)+[*]f”(ξ)(x-x0)2≥f(x0)+f’(x0)(x-x0) 将x=u(t)代入,并两边对t从0到a积分,有 ∫0af[u(t)]dt≥∫0af(x0)dt+∫0af’(x0)[u(t)-x0]dt 即 ∫0af[u(t)]dt≥af(x0)+f’(x0)[∫0au(t)dt-x0a] 令x0=[*]∫0au(t)dt,得 ∫0af[u(t)]dt≥af[[*]∫0au(t)dt] 即 [*]∫0af[u(t)]dt≥f[[*]∫0au(t)dt]

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