Question

Cho \(x>0;\) \(y>0;\) \(x+y\le1\). CM: \(\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}\ge4\)

Answer 1

\(VT\ge\dfrac{4}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}\ge4\) (vì \(x+y\le1\) )

Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)

Ta có đpcm