解答题 11.设f(x)在[0,1]上连续且单调减少,且f(x)>0.证明:
【正确答案】等价于∫01f2(x)dx∫01xf(x)dx≥∫01f(x)dx∫01xf2(x)dx,
等价于∫01f2(x)dx∫01yf(y)dy≥∫01f(x)dx∫01yf2(y)dy,或者
01dx∫01yf(x)f(y)[f(x)一f(y)]dy≥0
令I=∫01dx∫01yf(x)f(y)[f(x)一f(y)]dy,
根据对称性I=∫01dx∫01xf(x)y(y)[f(y)一f(x)]dy,
2I=∫01dx∫01f(x)f(y)(y—x)[f(x)一f(y)]dy,
因为f(x)>0且单调减少,所以(y—x)[f(x)一f(y)]≥0,于是2I≥0,或I≥0,
所以
【答案解析】