$1)\, y = 2\sin4x + 3\cos4x + 1$
Áp dụng bất đẳng thức $Bunyakovsky$ ta được:
$(y - 1)^2 = (2\sin4x + 3\cos4x)^2 \leq (2^2 + 3^2)(\sin^24x + \cos^24x) = 13$
$\Rightarrow -\sqrt{13} \leq y - 1 \leq \sqrt{13}$
$\Rightarrow 1 - \sqrt{13} \leq y \leq 1 + \sqrt{13}$
Vậy $\min y = 1 - \sqrt{13} \Leftrightarrow 2\sin4x + 4\cos4x = - \sqrt{13} \Leftrightarrow x = -\dfrac{\pi}{8} - \dfrac{1}{4}\arccos\dfrac{2}{\sqrt{13}} + k\dfrac{\pi}{2}$
$\max y = 1 + \sqrt{13} \Leftrightarrow 2\sin4x + 4\cos4x = \sqrt{13} \Leftrightarrow x = \dfrac{\pi}{8} - \dfrac{1}{4}\arccos\dfrac{2}{\sqrt{13}} + k\dfrac{\pi}{2} \qquad (k \in \Bbb Z)$
$2)\, y = \sin x + 2\sqrt{2 - \sin^2x}$
Ta có:
$\sin x \geq - 1$
$\sin^2x \geq 0$
$\Leftrightarrow 2 - \sin^2x \geq 1$
$\Leftrightarrow 2\sqrt{2 - \sin^2x} \geq 2$
$\Rightarrow \sin x + 2\sqrt{2 - \sin^2x} \geq -1 + 2 = 1$
Hay $y \geq 1$
Dấu = xảy ra $\Leftrightarrow \sin x = -1 \Leftrightarrow x = - \dfrac{\pi}{2} + k2\pi$
Áp dụng bất đẳng thức $Bunyakovsky$ ta được:
$y^2 = (\sin x+ 2\sqrt{2 - \sin^2x})^2 \leq (1^2 + 2^2)(\sin^2x + 2 - \sin^2x) = 10$
$\Rightarrow y \leq \sqrt{10}$
Dấu = xảy ra $\Leftrightarrow 2\sin x = \sqrt{2 - \sin^2x} \Leftrightarrow \sin x = \pm \sqrt{\dfrac{2}{5}}$
