c/m bpt lượng giác

K

kakashi_hatake

$sin^4x. cos^6x=2^2.(\dfrac{sin^2x}{2})^2.3^3.(\dfrac{cos^2x}{3})^3 \le 108.(\dfrac{2\dfrac{sin^2x}{2}+3\dfrac{cos^2x}{3}}{5})^5=\dfrac{108}{3125}$
Dấu bằng xảy ra khi $\dfrac{sin^2x}{2}=\dfrac{cos^2x}{3}=\dfrac{1}{5}$
 
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T

thien0526

Mình giải cách khác xem nhé:
Áp dụng BĐT Côsi cho 5 số k âm: [TEX]\frac{sin^2 x}{2}; \frac{sin^2 x}{2}; \frac{cos^2 x}{3}; \frac{cos^2 x}{3}; \frac{cos^2 x}{3}[/TEX], ta có
[TEX]\frac{\frac{sin^2 x}{2}+\frac{sin^2 x}{2}+\frac{cos^2 x}{3}+\frac{cos^2 x}{3}+\frac{cos^2 x}{3}}{5} \geq 2\sqrt[5]{\frac{sin^2 x}{2}.\frac{sin^2 x}{2}.\frac{cos^2 x}{3}.\frac{cos^2 x}{3}.\frac{cos^2 x}{3}}[/TEX]
[TEX]\Leftrightarrow\frac{1}{5}\geq\sqrt[5]{\frac{sin^4 x.cos^6 x}{108}}[/TEX]
[TEX]\Leftrightarrow sin^4 x.cos^6 x\leq\frac{108}{3125}[/TEX](ĐPCM)
Dấu "=" xảy ra thì như bạn ở trên nói rồi:D
 
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