Ta có: \(\left(x-y\right)^2\ge0\)
\(\Leftrightarrow x^2+y^2-2xy\ge0\)
\(\Leftrightarrow x^2+y^2\ge2xy\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
\(\Rightarrow A=\dfrac{x^2+y^2}{\left(x+y\right)^2}\ge\dfrac{x^2+y^2}{2\left(x^2+y^2\right)}=\dfrac{1}{2}\)
Dấu \("="\) xảy ra \(\Leftrightarrow x=y\) .