Question

[Nguyễn Thị Huyền Diệp]

Cho x,y>0. CMR: \(\left(x+y\right)^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{x^2+y^2}\right)\ge10\)

Answer 1

[Nguyễn Việt Lâm]

\(P=\left(x^2+y^2+2xy\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{x^2+y^2+2xy}{x^2+y^2}\)

\(P=\left(x^2+y^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2xy\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+1+\dfrac{2xy}{x^2+y^2}\)

\(P\ge2xy.\dfrac{2}{xy}+\dfrac{2\left(x^2+y^2\right)}{xy}+1+\dfrac{2xy}{x^2+y^2}\)

\(P\ge\dfrac{x^2+y^2}{2xy}+\dfrac{2xy}{x^2+y^2}+\dfrac{3}{2}\left(\dfrac{x^2+y^2}{xy}\right)+5\)

\(P\ge2\sqrt{\dfrac{2xy\left(x^2+y^2\right)}{2xy\left(x^2+y^2\right)}}+\dfrac{3}{2}.\dfrac{2xy}{xy}+5=10\)

Dấu "=" xảy ra khi \(x=y\)