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Question

Tìm giá trị lớn nhất của biểu thức

$A=\alpha\left(\sin^2A+\sin^2B+\sin^2C\right)-\beta\left(\cos^3A+\cos^3B+\cos^3C\right)$ trong đó

$A;\,B;\,C$ là độ lớn ba góc của một tam giác nhọn và

$\alpha;\,\beta$ là hai số dương cho trước.

khangnguyenthanh · Answer 1 · 7/27/2013 8:22:37 AM

Ta có:

$\sin^2A+\sin^2B+\sin^2C$

$=2-\dfrac{1}{2}(\cos2A+\cos2B)-\cos^2C$

$=2-\cos(A+B)\cos(A-B)-\cos^2C$

$=2+\cos C[\cos(A-B)-\cos C]$

$\le2+\cos C(1-\cos C)$

$\le2+\left[\dfrac{\cos C+(1-\cos C)}{2}\right]^2=\dfrac{9}{4}$

Từ đó suy ra:

$\cos^2A+\cos^2B+\cos^2C\ge\dfrac{3}{4}$

Áp dụng BĐT Cauchy ta có:

$\cos^3A+\cos^3A+\dfrac{1}{8}\ge\dfrac{3}{2}\cos^2A$

$\cos^3B+\cos^3B+\dfrac{1}{8}\ge\dfrac{3}{2}\cos^2B$

$\cos^3B+\cos^3B+\dfrac{1}{8}\ge\dfrac{3}{2}\cos^2C$

Suy ra:

$\cos^3A+\cos^3B+\cos^3C\ge\dfrac{3}{8}$

$\Rightarrow A\le\dfrac{9\alpha}{4}-\dfrac{3\beta}{8}$

Vậy:

$\max A=\dfrac{9\alpha}{4}-\dfrac{3\beta}{8} \Leftrightarrow A=B=C=\dfrac{\pi}{3}$

Trả lời 27-07-13 08:22 AM

khangnguyenthanh
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chỗ >= 3/8 là sao vay bn, minh khong hiu cho do – minhthanh2105 27-07-13 01:14 PM

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