$\cos3x + \sqrt{2 - \cos^{2} 3x} = 2( 1+ \sin^{2} 2x)$
Áp dụng BĐT BCS cho vế trái, ta có:
$cos3x+\sqrt{2-cos^23x}\leq \sqrt{(1^2+1^2)(cos^23x+2-cos^23x)}=2$
Mặt khác: $VP=2(1+sin^22x)\geq 2$
Dấu "=" xảy ra khi $\begin{cases}sin2x=0 \\ cos3x+\sqrt{2-cos^23x}=2 \end{cases}$
ê nhưng mà dấu = xảy ra khi cos3x = căn 2-cos^2 3x chứ –  ngoinhacuameo123 02-12-14 10:03 PM
ok hiểu rồi . tiện thể xem hộ mình bài này thấy chả sai chỗ nào mà không ra đáp án 1/sinx = 1/sin2x 1/sin4x=0<=> 1/sinx(1-1/2cosx -1/4cosxcos2x)=0<=> 1/sinx(4cosxcos2x -2cos2x -1)=0<=> 4cosx(2cos^2 x -1) -2(2cos^2 x -1)-1=0 –  ngoinhacuameo123 02-12-14 09:48 PM
Bunhiacopxki –  Nero 02-12-14 09:39 PM
bất đẳng thức BCS là gì thế –  ngoinhacuameo123 02-12-14 09:34 PM

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