\(x^2+y^2\le x+y\Leftrightarrow\left(2x-1\right)^2\le-4y^2+4y+1\text{ (1)}\)
+Nếu \(-4y^2+4y+1< 0\) thì (1) có \(VT\ge0>VP\), (1) ko thỏa --> loại.
+Nếu \(-4y^2+4y+1=0\Leftrightarrow y=\frac{1+\sqrt{2}}{2}\text{ }\left(do\text{ }y>0\right)\) thì\(\left(2x-1\right)^2\le0\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)
\(A=x+3y=2+\frac{3}{\sqrt{2}}\approx4.12\)
+Xét \(-4y^2+4y+1>0\Leftrightarrow\frac{1-\sqrt{2}}{2}< y< \frac{1+\sqrt{2}}{2}\)
\(\Rightarrow0< y< \frac{1+\sqrt{2}}{2}\approx1.207\)
\(\left(1\right)\Leftrightarrow-\sqrt{-4y^2+4y+1}\le2x-1\le\sqrt{-4y^2+4y+1}\)
\(\Rightarrow2x\le\sqrt{2-\left(2y-1\right)^2}+1\)
\(2A=2x+6y\le\sqrt{2-\left(2y-1\right)^2}+3\left(2y-1\right)+1+3\)
Áp dụng bđt Bu-nhia-cop-xki
\(1.\sqrt{2-\left(2y-1\right)^2}+3.\left(2y-1\right)\le\sqrt{1^2+3^2}.\sqrt{2-\left(2y-1\right)^2+\left(2y-1\right)^2}=2\sqrt{5}\)
Dấu bằng xảy ra khi \(\frac{1}{3^2}=\frac{2-\left(2y-1\right)^2}{\left(2y-1\right)^2}\Leftrightarrow\left(2y-1\right)^2=\frac{9}{5}\)
\(\Leftrightarrow2y-1=\pm\frac{3}{\sqrt{5}}\Leftrightarrow\orbr{\begin{cases}y=\frac{3}{2\sqrt{5}}+\frac{1}{2}\approx1.17\in\left(0;\frac{1+\sqrt{2}}{2}\right)\\y=-\frac{3}{2\sqrt{5}}+\frac{1}{2}< 0\end{cases}}\)
\(\Rightarrow2A\le4+2\sqrt{5}\)
\(\Rightarrow A\le2+\sqrt{5}\approx4.23\)
Dấu bằng xảy ra khi \(\hept{\begin{cases}y=\frac{3}{2\sqrt{5}}+\frac{1}{2}\\x=\frac{1+\sqrt{2-\left(2y-1\right)^2}}{2}=\frac{1}{2\sqrt{5}}+\frac{1}{2}\end{cases}}\)