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b. Bổ đề: $\sin x>\frac{2x}{\pi},\forall x\in\left(0,\frac{\pi}{2}\right)$.
Thật vậy:
Xét hàm: $f(x)=\frac{\sin x}{x}, x\in\left(0,\frac{\pi}{2}\right) $
Ta có: $f'(x)=\frac{x-\tan x}{x^2\cos x}$
Xét hàm: $g(x)=x-\tan x, x\in\left(0,\frac{\pi}{2}\right) $
Ta có: $g'(x)=-\cos^2x<0, \forall x\in\left(0,\frac{\pi}{2}\right) $
$\Rightarrow g(x)<g(0)=0,\forall x\in\left(0,\frac{\pi}{2}\right) $
Hay $f'(x)<0 \forall x\in\left(0,\frac{\pi}{2}\right) $
$\Rightarrow f(x)>f(\frac{\pi}{2})=\frac{2}{\pi},
\forall x\in\left(0,\frac{\pi}{2}\right)$ , bổ đề được chứng minh.
Xét hàm: $h(x)=\cos x-1+\frac{x^2}{\pi},
x\in\left(0,\frac{\pi}{2}\right)$
Ta có: $h'(x)=-\sin x+\frac{2x}{\pi}<0,
\forall x\in\left(0,\frac{\pi}{2}\right) $
$\Rightarrow h(x)>h(0)=0$, đpcm.
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