Lời giải
Đề bài:
Chứng minh rằng:$\left ( \sin x +a\cos x\right )\left ( \sin x +b\cos x\right )\leq \frac{1}{2}[1+ab+\sqrt{\left ( 1+a^{2} \right )\left ( 1+b^{2} \right )}]$
Lời giải
Ta có: $\left ( \sin x +a\cos x\right )\left ( \sin x +b\cos x\right )$
$= \sin ^{2}x+\left ( a+b \right )\sin x .\cos x+ab\cos ^{2}x$
$=\frac{1}{2}\left (1-\cos 2x\right )+\frac{1}{2}\left ( a+b \right )\sin 2x+\frac{1}{2}ab\left (1+\cos 2x\right )$
$=\frac{1}{2}\left ( 1+ab \right )+\frac{1}{2}\left ( a+b \right )\sin 2x+\frac{1}{2}\left ( ab-1 \right )\cos 2x$
$\leq \frac{1}{2}\left ( 1+ab \right )+\frac{1}{2}\sqrt{\sin ^{2}2x+\cos ^{2}2x}.\sqrt{\left ( a+b \right )^{2}+\left ( ab-1 \right )^{2}}$( theo BĐT Bunhiacopxki)
$\leq \frac{1}{2}\left ( 1+ab \right )+\frac{1}{2}\sqrt{a^{2}+b^{2}+a^{2}b^{2}+1}$
$\leq \frac{1}{2}\left ( 1+ab+\sqrt{\left ( 1+a^{2} \right )\left ( 1+b^{2} \right )}\right )\Rightarrow $ (ĐPCM)
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Chuyên mục: Bất đẳng thức lượng giác
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