Ta có \(x;y\) phải không âm thì mới xảy ra \(\sqrt{x}+2\sqrt{y}=2\)
Áp dụng BĐT bunhiacopxki ta có:
\(4=\left(\sqrt{x}+2\sqrt{y}\right)^2\le\left(1^2+2^2\right)\left(x+y\right)\)\(\Rightarrow x+y\ge\dfrac{4}{5}\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{y}=2\\x;y\ge0\\\dfrac{x}{1}=\dfrac{y}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2\sqrt{2}+1\right)\sqrt{x}=2\\x;y\ge0\\y=2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{\left(2\sqrt{2}+1\right)^2}=\dfrac{4\left(2\sqrt{2}-1\right)^2}{49}\\y=\dfrac{8}{\left(2\sqrt{2}+1\right)^2}=\dfrac{8\left(2\sqrt{2}-1\right)^2}{49}\end{matrix}\right.\)
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