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变型∫(0,λ)f(x)dx >= λ(∫(0,λ)f(x)dx + ∫(λ,1)f(x)dx)(1-λ)∫(0,λ)f(x)dx >= λ∫(λ,1)f(x)dx即只需证 [∫(0,λ)f(x)dx]/λ >= [∫(λ,1)f(x)dx]/(1-λ)因为f(x)在[0,1]上单调减少，则在[0,λ]上有f(x)>=f(λ),在[λ,1]上有f(x)= ∫(0,λ)f(λ)dx = f(λ)λ,∫(λ,1)f(x)dx >= ∫(λ,1)f(λ)dx = f(λ)(1-λ),所以 [∫(0,λ)f(x)dx]/λ >= [∫(λ,1)f(x)dx]/(1-λ),所以原不等式成... |
