[Everyday Mathematics]20150127
2015-01-13 09:43
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设 $f,g:[a,b]\to [0,\infty)$ 连续, 单调递增, 并且 $$\bex \int_a^x \sqrt{f(t)}\rd t\leq \int_a^x \sqrt{g(t)}\rd t,\quad \forall\ x\in [a,b];\quad\quad\int_a^b \sqrt{f(t)}\rd t= \int_a^b \sqrt{g(t)}\rd t. \eex$$ 试证: $$\bex \int_a^b \sqrt{1+f(t)}\rd t\leq \int_a^b \sqrt{1+g(t)}\rd t. \eex$$
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