\(\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}\)
\(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\)
Ta có:
\(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{4}{\left(x+y\right)^2}\)
\(xy\le\dfrac{\left(x+y\right)^2}{4}\)
\(\Rightarrow2xy\le\dfrac{\left(x+y\right)^2}{2}\)
\(\Rightarrow\dfrac{1}{2xy}\ge\dfrac{2}{\left(x+y\right)^2}\)
\(\Rightarrow\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{2}{\left(x+y\right)^2}=\dfrac{6}{\left(x+y\right)^2}\)
phải cho x+y=1 nha bạn
Đúng 0
Bình luận (0)
