Question

2.Cho x,y>0 thỏa mãn x+y=1.Tìm min của A=\(\frac{1}{x^2+y^2}\)+\(\frac{2}{xy}\) +4xy

Answer 1

\(A=\frac{1}{x^2+y^2}+\frac{2}{xy}+4xy=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(4xy+\frac{1}{4xy}\right)+\frac{5}{4xy}\)

\(\ge\frac{\left(1+1\right)^2}{x^2+2xy+y^2}+2+\frac{5}{\left(x+y\right)^2}=4+2+5=11\)

Answer 2

A = \(\frac{7}{2}\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(\frac{1}{4xy}+4xy\right)-\frac{5}{2\left(x^2+y^2\right)}\)

Áp dụng bđt cauchy là ra bài