\(M=4xy+\frac{1}{x^2+y^2}+\frac{2}{xy}\)
\(=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{3}{2xy}+24xy-20xy\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel, ta có:
\(\frac{1}{x^2+y^2}+\frac{1}{2xy}\ge\frac{4}{\left(x+y\right)^2}\)
Áp dụng BĐT AM-GM, ta có:
\(\frac{3}{2xy}+24xy\ge12\)
Và: \(1\ge x+y\)
\(\Leftrightarrow1\ge\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow-20xy\ge-5\)
\(\Rightarrow M\ge4+12-5=11\)
\(''=''\Leftrightarrow x=y=\frac{1}{2}\)
Vậy...
\(M=\frac{1}{x^2+y^2}+\frac{1}{2xy}+4xy+\frac{1}{4xy}+\frac{5}{4xy}\)
\(M\ge\frac{4}{x^2+y^2+2xy}+2\sqrt{\frac{4xy}{4xy}}+\frac{5}{\left(x+y\right)^2}\)
\(M\ge\frac{4}{\left(x+y\right)^2}+\frac{5}{\left(x+y\right)^2}+2=\frac{9}{\left(x+y\right)^2}+2\ge\frac{9}{1}+2=11\)
\(\Rightarrow M_{min}=11\) khi \(x=y=\frac{1}{2}\)
