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a) Ta có : $ \cos x < 1 , \forall x \in \left ( 0;\frac{\pi}{2} \right ] \Rightarrow \int\limits_{0}^{t}\cos xdx < \int\limits_{0}^{t}dt, \forall t \in \left ( 0;\frac{\pi}{2} \right ] $
$\Rightarrow \sin t < t, \forall t \in \left ( 0;\frac{\pi}{2} \right ] \Rightarrow \int\limits_{0}^{x} \sin tdt < \int\limits_{0}^{x}tdt, \forall t \in \left ( 0;\frac{\pi}{2} \right ] $
$\Rightarrow 1 - \cos x < \frac{x^2}{2} , \forall x \in \left ( 0;\frac{\pi}{2} \right ] $
$\Rightarrow \int\limits_{0}^{t} ( 1 - \cos x)dx < \int\limits_{0}^{t} \frac{x^2}{2}dx , \forall t \in \left ( 0;\frac{\pi}{2} \right ] $
$\Rightarrow t - \sin t < \frac{t^3}{6}, \forall t \in \left ( 0;\frac{\pi}{2} \right ] \Rightarrow \sin t > t -\frac{t^3}{6}, \forall t \in \left ( 0;\frac{\pi}{2} \right ]$ (1)
$\Rightarrow \int\limits_{0}^{x}\sin tdt > \int\limits_{0}^{x} \left ( t-\frac{t^3}{6} \right )dt, \forall x \in \left ( 0;\frac{\pi}{2} \right ]$
$\Rightarrow 1 - \cos x > \frac{x^2}{2} - \frac{x^4}{24} , \forall x \in \left ( 0;\frac{\pi}{2} \right ]$
$\Rightarrow \cos x < 1 -\frac{x^2}{2} +\frac{x^4}{24}, \forall x \in \left ( 0;\frac{\pi}{2} \right ]$ (2)
Từ (1) và (2) $ \Rightarrow \left ( \frac{\sin x}{x} \right ) ^3 > \left ( 1 - \frac{x^2}{6} \right )^3 \geq 1 - \frac{x^2}{2}+\frac{x^4}{12}-\frac{x^6}{216}$
$ > 1 - \frac{x^2}{2} +\frac{x^4}{24} > \cos x , \forall x \in \left ( 0;\frac{\pi}{2} \right ]$
Vậy : $ \left ( \frac{\sin x}{x} \right )^3 > \cos x, \forall x \in \left ( 0;\frac{\pi}{2} \right ]$ (3)
b) Từ (3) $ \Rightarrow \frac{\cos x }{\sin ^3 x} < \frac{1}{x^3} , \forall x \in \left ( 0;\frac{\pi}{2} \right ] $
$\Rightarrow \int\limits_{x}^{\frac{\pi}{2} } \frac{\cos t}{\sin ^3t}dt \leq \int\limits_{x}^{\frac{\pi}{2} } \frac{dt}{t^3}, \forall x \in \left ( 0;\frac{\pi}{2} \right ] $
$\Rightarrow \frac{1}{\sin ^2 x} - \frac{1}{x^2} \leq 1 - \frac{4}{\pi ^2}, \forall x \in \left ( 0;\frac{\pi}{2} \right ]$
Dấu $"=" \Leftrightarrow x = \frac{\pi}{2} \Rightarrow P \leq 3 - \frac{12}{\pi^2 }, \forall x ,y,z \in \left ( 0;\frac{\pi}{2} \right ]$
Dấu $"=" \Leftrightarrow x = y = z = \frac{\pi}{2}$
vậy $Max P = 3 - \frac{12}{\pi ^2}.$
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