\(A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{4}{\left(x+y\right)^2}+\frac{1}{2xy}\\ =\frac{1}{4}+\frac{1}{2xy}\ge\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\)
Dấu = xảy ra khi x=y=2
vay la ??????????????????????????
ta có: \(A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)
Áp dụng BĐT AM-GM ta có\(\frac{1}{x^2+y^2}+\frac{1}{2xy}\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}=\frac{2^2}{\left(x+y\right)^2}=\frac{4}{4^2}=\frac{1}{4}\)
Áp dụng BĐT Cô-si ta có:\(x+y\ge2\sqrt{xy}\Rightarrow\left(x+y\right)^2\ge4xy\Rightarrow xy\le\frac{\left(x+y\right)^2}{4}\)(1)
\(\Rightarrow2xy\le\frac{\left(x+y\right)^2}{2}\Rightarrow\frac{1}{2xy}\ge\frac{2}{\left(x+y\right)^2}=\frac{2}{4^2}=\frac{1}{8}\)(2)
Từ (1), (2)\(\Rightarrow A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\)
Vậy\(Amin=\frac{3}{8}\)đạt được khi\(x=y=2\)
(1) là ở phía sau 1/4 nha, cho mình xin lỗi
chao tra mi minh la long ban cua cau va
