[2003年]设函数f(x)连续且恒大于零, 其中Ω(t)={(x,y,z)|x2+y2+z2≤t2},D(t)={(x,y)|x2+y2≤t2}. 证明当t>0时,F(t)>(2/π)G(t).

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问题 [2003年]设函数f(x)连续且恒大于零,

其中Ω(t)={(x,y,z)|x2+y2+z2≤t2},D(t)={(x,y)|x2+y2≤t2}.
证明当t>0时,F(t)>(2/π)G(t).

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答案F(t)一[*]=2{∫0tf(r2)r2dr∫0tf(r2)dr—[∫0tf(r2)rdr]2}/[∫01f(r2)dr∫01rf(r2)dr]. 令g(t)=∫0tf(r2)r2dr∫0tf(r2)dr一[∫0trf(r2)dr]2,则g(0)=0.又因f(x)恒大于零,有 g’(t)=f(t2)t20tf(r2)dr+f(t2)∫0tf(r2)r2dr一2f(t2)t∫0tf(r2)rdr =f(t2)[∫0tf(t2)t2dr+∫0tf(r2)r2dr—2∫0tf(r2)rtdr] =f(t2)∫0tf(t2)(t2一2rt+r2)dr =f(t2)∫0tf(t2)(t一r)2dr>0. 故g(t)在(0,+∞)内单调增加,又g(0)=0,所以当t>0时有g(t)>0,又∫0tf(r2)dr∫01rf(r2)dr>0,故当t>0时 [*] 即[*]

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