\(A=\dfrac{1}{4}.4ab\left(a^2+b^2-2ab\right)\le\dfrac{1}{16}\left(4ab+a^2+b^2-2ab\right)=\dfrac{1}{16}\left(a+b\right)^2=\dfrac{1}{16}\)
Dấu "=" xảy ra khi \(\left(a;b\right)=\left(\dfrac{2-\sqrt{2}}{4};\dfrac{2+\sqrt{2}}{4}\right);\left(\dfrac{2+\sqrt{2}}{4};\dfrac{2-\sqrt{2}}{4}\right)\)