Question

Cho x² + y² = x + y . Tìm Max F = xy

Answer 1

\(x^2+y^2=x+y\Rightarrow x+y>0\)

Áp dụng bất đẳng thức: \(\left(a-b\right)^2\ge0\Leftrightarrow ab\le\frac{\left(a+b\right)^2}{4}\)

\(\left(x+y\right)^2-\left(x+y\right)=2xy\le\frac{1}{2}\left(x+y\right)^2\)

\(\Rightarrow\left(x+y\right)^2-2\left(x+y\right)\le0\)

\(\Leftrightarrow\left(x+y\right)\left(x+y-2\right)\le0\)

\(\Leftrightarrow x+y\le2\text{ }\left(do\text{ }x+y\ge0\right)\)

Do đó: \(4\ge\left(x+y\right)^2\ge4xy\)

\(\Rightarrow xy\le1\)

Đẳng thức xảy ra khi \(x=y;\text{ }x+y=2\Leftrightarrow x=y=1.\)

Vậy Max F = 1.