Ta có:
\(3sin^4x+cos^4x=\frac{\left(sin^2x\right)^2}{\frac{1}{3}}+\frac{\left(cos^2x\right)^2}{1}\ge\frac{\left(sin^2x+cos^2x\right)^2}{\frac{1}{3}+1}=\frac{1}{\frac{4}{3}}=\frac{3}{4}\)
Dấu "=" xảy ra khi và chỉ khi \(3sin^2x=cos^2x\Leftrightarrow4sin^2x=1\Rightarrow sin^2x=\frac{1}{4}\Rightarrow cos^2x=\frac{3}{4}\)
\(\Rightarrow A=\left(\frac{1}{4}\right)^2+3.\left(\frac{3}{4}\right)^2=\frac{7}{4}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\((3\sin ^4x+\cos ^4x)(\frac{1}{3}+1)\geq (\sin ^2x+\cos ^2x)^2=1\)
\(\Leftrightarrow 3\sin ^4x+\cos ^4x\geq \frac{3}{4}\)
Dấu "=" xảy ra khi \(3\sin ^2x=\cos ^2x\). Mà $\sin ^2x+\cos ^2x=1$ nên suy ra:
$\sin ^2x=\frac{1}{4}; \cos ^2x=\frac{3}{4}$
$\Rightarrow A=(\frac{1}{4})^2+3(\frac{3}{4})^2=\frac{7}{4}$
