关于不等式的专题讨论II(构造似序不等式并积分的思想，重积分的思想，构造变上限积分思想与单调性的应用)

$f命题：$ $(f{似序不等式})$设$fleft( x ight),gleft( x ight)$均在$left[ {a,b} ight]$上连续且函数的单调性相同，则

[left( {b - a} ight)int_a^b {fleft( x ight)gleft( x ight)dx}  ge int_a^b {fleft( x ight)dx} int_a^b {gleft( x ight)dx} ]

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$f命题：$$(f{Tschebyscheff不等式})$设$pleft( x ight)$在$left[ {a,b} ight]$非负连续，$fleft( x ight),gleft( x ight)$在$left[ {a,b} ight]$上连续且单调递增，则

[left( {int_a^b {pleft( x ight)fleft( x ight)dx} } ight)left( {int_a^b {pleft( x ight)gleft( x ight)dx} } ight) le left( {int_a^b {pleft( x ight)dx} } ight)left( {int_a^b {pleft( x ight)fleft( x ight)gleft( x ight)dx} } ight)]

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$f命题：$$(f{Kantorovich不等式})$设正值函数$fleft( x ight)$在$left[ {0,1} ight]$上连续，$m = mathop {min}limits_{x in left[ {0,1} ight]} fleft( x ight),M = mathop {max}limits_{x in left[ {0,1} ight]} fleft( x ight)$，则

[int_0^1 {fleft( x ight)dx} int_0^1 {frac{1}{{fleft( x ight)}}dx}  le frac{{{{left( {m + M} ight)}^2}}}{{4mM}}]

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$f命题：$设$f:[-1,1] o { ext{R}}$为偶函数，$f$在$[0,1]$上单调递增，又设$g$是$[-1,1]$上的凸函数，即对任意的$x,yin [-1,1]$及$tin (0,1)$有$gleft( {tx + left( {1 - t} ight)y} ight) leqslant tgleft( x ight) + left( {1 - t} ight)gleft( y ight)$，则[2int_{ - 1}^1 {fleft( x ight)gleft( x ight)dx}  geqslant int_{ - 1}^1 {fleft( x ight)dx} int_{ - 1}^1 {gleft( x ight)dx} ]

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$f命题：$设$fleft( x ight)$在$left[ {a,b} ight]$连续且单调递增，则

[int_a^b {xfleft( x ight)dx}  ge frac{{a + b}}{2}int_a^b {fleft( x ight)dx} ]

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$f命题：$设函数$fleft( x ight)$在$left[ {0,1} ight]$上连续，且$0 le fleft( x ight) < 1$，则[int_0^1 {frac{{fleft( x ight)}}{{1 - fleft( x ight)}}dx ge frac{{int_0^1 {fleft( x ight)dx} }}{{1 - int_0^1 {fleft( x ight)dx} }}} ]

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$f命题：$设$a,b > 0,fleft( x ight) geqslant 0,fleft( x ight) in Rleft[ {a,b} ight],int_a^b {xfleft( x ight)dx}  = 0$，则[int_a^b {{x^2}fleft( x ight)dx}  leqslant abint_a^b {fleft( x ight)dx} ]

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$f命题：$设$fin (0,1)$，且无穷积分$int_0^{ + infty } {fleft( x ight)dx} ,int_0^{ + infty } {xfleft( x ight)dx} $均收敛，证明：[int_0^{ + infty } {xfleft( x ight)dx}  > frac{1}{2}{left( {int_0^{ + infty } {fleft( x ight)dx} } ight)^2}]

$f命题：$设$f(x)$为$[0,+infty)$上非负可导函数，且$fleft( 0 ight) = 0,f'left( x ight) leqslant frac{1}{2}$，若$int_0^{ + infty } {fleft( x ight)dx} $收敛，证明：对$forall alpha  > 1$，有$int_0^{ + infty } {{f^alpha }left( x ight)dx} $收敛，且[int_0^{ + infty } {{f^alpha }left( x ight)dx}  leqslant {left( {int_0^{ + infty } {fleft( x ight)dx} } ight)^eta },eta  = frac{{alpha  + 1}}{2}]

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$f命题：$$(f{Bellman -Gronwall不等式})$设常数$k > 0$，函数$f,g$在$left[ {a,b} ight]$上非负连续，且对任意$x in left[ {a,b} ight]$满足

[fleft( x ight) ge k + int_a^x {gleft( t ight)fleft( t ight)dt} ]
证明：对任意$x in left[ {a,b} ight]$，有$fleft( x ight) ge k{e^{int_a^x {gleft( t ight)dt} }}$

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$f命题：$