$\bf命题：$设$f,g \in R\left[ {a,b} \right]$，且${m_1} \le f\left( x \right) \le {M_1},{m_2} \le g\left( x \right) \le {M_2}$，则
\[\frac{1}{{b - a}}\int_a^b {f\left( x \right)g\left( x \right)dx} - \frac{1}{{{{\left( {b - a} \right)}^2}}}\int_a^b {f\left( x \right)dx} \int_a^b {g\left( x \right)dx} \le \frac{{\left( {{M_1} - {m_1}} \right)\left( {{M_2} - {m_2}} \right)}}{4}\]
$\bf命题：$设连续函数$f\left( x \right)$$:\left[ {0,\infty } \right) \to \left[ {0,\infty } \right)$满足$f\left( {f\left( x \right)} \right) = {x^m},\forall x \in \left[ {0,\infty } \right),m \in {Z^ + }$，则\[\int_0^1 {{f^2}\left( x \right)dx} \ge \frac{{2m - 1}}{{{m^2} + 6m - 3}}\]
$\bf命题：$设$p > 1,f\left( x \right) \in C\left( {0, + \infty } \right),\int_0^{ + \infty } {{{\left| {f\left( t \right)} \right|}^p}dt} $收敛，证明：\[{\left\{ {\int_0^{ + \infty } {{{[\frac{1}{x}\int_0^x {\left| {f\left( t \right)dt} \right|} ]}^p}dx} } \right\}^{\frac{1}{p}}} \leqslant \frac{p}{{p - 1}}{\left( {\int_0^{ + \infty } {{{\left| {f\left( t \right)} \right|}^p}dt} } \right)^{\frac{1}{p}}}\]
$\bf命题：$
附录
$\bf命题：$设$f\left( x \right) \in {C^1}\left[ {a,b} \right]$，且$f\left( a \right) = 0$，则\[\int_a^b {\left| {f\left( x \right)f'\left( x \right)} \right|dx} \le \frac{{b - a}}{2}\int_a^b {{{\left[ {f'\left( x \right)} \right]}^2}dx} \]
当且仅当$f\left( x \right){\rm{ = }}k\left( {x - a} \right)$时等号成立
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$\bf命题：$设$f\left( x \right) \in {C^1}\left[ {a,b} \right]$，且$f\left( a \right) = f\left( b \right) = 0$，则\[\int_a^b {\left| {f\left( x \right)f'\left( x \right)} \right|dx} \le \frac{{b - a}}{4}\int_a^b {{{\left[ {f'\left( x \right)} \right]}^2}dx} \]
并且$\frac{{b - a}}{4}$为最佳系数
$\bf(04中科院七)$设$0<x,y<\pi$，求证：$\sin x\sin y\sin \left( {x + y} \right) \leqslant \frac{{3\sqrt 3 }}{8}$
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$\bf(对数不等式)$当$x>-1$时，成立：$\frac{x}{{1 + x}} \leqslant \ln \left( {1 + x} \right) \leqslant x$
$\bf(Jordan不等式)$当$0<x<\frac{\pi }{2}$时，成立：$\frac{2}{\pi }x \leqslant \sin x \leqslant x$