Đáp án: $y\le\sqrt{10}$
Giải thích các bước giải:
Ta có:
$y=\sqrt{1+2\cos^2x}+\dfrac12\sqrt{5+2\sin^2x}$
$\to y^2=(\sqrt{1+2\cos^2x}+\dfrac12\sqrt{5+2\sin^2x})^2$
$\to y^2\le (1^2+(\dfrac12)^2)(1+2\cos^2x+5+2\sin^2x)$ (BĐT Bunhiacopxki)
$\to y^2\le 10$
$\to y\le \sqrt{10}$
Dấu = xảy ra khi:
$\dfrac{\sqrt{1+2\cos^2x}}{1}=\dfrac{\sqrt{5+2\sin^2x}}{\dfrac12}$
$\to 2\sqrt{1+2\cos^2x}=\sqrt{5+2\sin^2x}$
$\to 4(1+\cos^2x)=5+2\sin^2x$
$\to 4+4\cos^2x=5+2(1-\cos^2x)$
$\to 4+4\cos^2x=7-2\cos^2x$
$\to 6\cos^2x=3$
$\to \cos^2x=\dfrac12$
$\to \cos x=\pm\sqrt{\dfrac12}$
$\to x\in\{\dfrac{\pi}{4}+k\pi\}$
