Ta có \(\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(x+y\right)^2-4xy\ge0\\ \Leftrightarrow\left(x+y\right)^2\ge4xy\\ \Leftrightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\\ \Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Dấu \("="\Leftrightarrow x=y\)
Áp dụng BĐT trên với BĐT cosi:
\(A=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{1}{4xy}+4xy+\dfrac{5}{4xy}\\ \Leftrightarrow A\ge\dfrac{4}{x^2+2xy+y^2}+2\sqrt{\dfrac{1}{4xy}\cdot4xy}+\dfrac{5}{\left(x+y\right)^2}\\ \Leftrightarrow A\ge\dfrac{4}{\left(x+y\right)^2}+\dfrac{5}{\left(x+y\right)^2}+2=\dfrac{4}{1}+\dfrac{5}{1}+2=11\)
Dấu \("="\Leftrightarrow x=y=\dfrac{1}{2}\)
