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)

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

\[\left( {b - a} \right)\int_a^b {f\left( x \right)g\left( x \right)dx} \ge \int_a^b {f\left( x \right)dx} \int_a^b {g\left( x \right)dx} \]

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$\bf命题：$$(\bf{Tschebyscheff不等式})$设$p\left( x \right)$在$\left[ {a,b} \right]$非负连续，$f\left( x \right),g\left( x \right)$在$\left[ {a,b} \right]$上连续且单调递增，则

\[\left( {\int_a^b {p\left( x \right)f\left( x \right)dx} } \right)\left( {\int_a^b {p\left( x \right)g\left( x \right)dx} } \right) \le \left( {\int_a^b {p\left( x \right)dx} } \right)\left( {\int_a^b {p\left( x \right)f\left( x \right)g\left( x \right)dx} } \right)\]

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

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

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$\bf命题：$设$f:[-1,1] \to {\text{R}}$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$x,y\in [-1,1]$及$t\in (0,1)$有$g\left( {tx + \left( {1 - t} \right)y} \right) \leqslant tg\left( x \right) + \left( {1 - t} \right)g\left( y \right)$，则\[2\int_{ - 1}^1 {f\left( x \right)g\left( x \right)dx} \geqslant \int_{ - 1}^1 {f\left( x \right)dx} \int_{ - 1}^1 {g\left( x \right)dx} \]

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

\[\int_a^b {xf\left( x \right)dx} \ge \frac{{a + b}}{2}\int_a^b {f\left( x \right)dx} \]

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

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$\bf命题：$设$a,b > 0,f\left( x \right) \geqslant 0,f\left( x \right) \in R\left[ {a,b} \right],\int_a^b {xf\left( x \right)dx} = 0$，则\[\int_a^b {{x^2}f\left( x \right)dx} \leqslant ab\int_a^b {f\left( x \right)dx} \]

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$\bf命题：$设$f\in (0,1)$，且无穷积分$\int_0^{ + \infty } {f\left( x \right)dx} ,\int_0^{ + \infty } {xf\left( x \right)dx} $均收敛，证明：\[\int_0^{ + \infty } {xf\left( x \right)dx} > \frac{1}{2}{\left( {\int_0^{ + \infty } {f\left( x \right)dx} } \right)^2}\]

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

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

\[f\left( x \right) \ge k + \int_a^x {g\left( t \right)f\left( t \right)dt} \]
证明：对任意$x \in \left[ {a,b} \right]$，有$f\left( x \right) \ge k{e^{\int_a^x {g\left( t \right)dt} }}$

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$\bf命题：$