Áp dụng bđt Bunhiacopxki , ta có : \(4=\left(x+y\right)^2=\left(\frac{1}{\sqrt{3}}.\sqrt{3}.x+1.y\right)^2\le\left[\left(\frac{1}{\sqrt{3}}\right)^2+1^2\right].\left(3x^2+y^2\right)\)
\(\Rightarrow3x^2+y^2\ge\frac{4}{\frac{1}{3}+1}=3\) \(\Rightarrow A\ge3\)
Vậy Min A = 3 \(\Leftrightarrow\hept{\begin{cases}x+y=2\\3x=y\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{2}\end{cases}}\)
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