\(Ta \) \(có : x^2 +y^2 +xy = 1\)

\(\Leftrightarrow\)\(xy = 1 - x^2 - y^2\)

\(Thay \)  \(xy = 1 - x^2 - y^2 \)  \(vào \)  \(P , ta \) \(được :\)

\(P = 1 - x^2 -y^2\)

\(P = 1 - ( x^2 +y^2 )\)

\(P = - ( x^2 +y^2 )+ 1\)\(\le\)\(1\)

\(Dấu "=" xảy \) \(ra\)  \(\Leftrightarrow\)\(x^2+y^2 =0\)

\(\Leftrightarrow\)\(x = 0 \) \(và\)  \(y = 0\)

\(Max \)  \(P = 1 \)\(\Leftrightarrow\)\(x = 0 ; y = 0\)

\(x+y=4xy\Rightarrow\frac{x+y}{xy}=\frac{1}{x}+\frac{1}{y}=4\)

\(\frac{1}{x}+\frac{1}{y}>=\frac{4}{x+y}\Rightarrow4>=\frac{4}{x+y}\Rightarrow x+y>=1\)(bđt svacxo)

\(x^2+y^2>=\frac{\left(x+y\right)^2}{2};xy< =\frac{\left(x+y\right)^2}{4}\)

\(\Rightarrow P=x^2+y^2-xy>=\frac{\left(x+y\right)^2}{2}-\frac{\left(x+y\right)^2}{4}=\frac{\left(x+y\right)^2}{4}>=\frac{1^2}{4}=\frac{1}{4}\)

dấu = xảy ra khi \(x+y=1;x=y\Rightarrow x=y=\frac{1}{2}\left(tm\right)\)

vậy min P là \(\frac{1}{4}\)khi x=y=\(\frac{1}{2}\)