Question

Cho x,y > 0 và thỏa mãn x² + y² = 2

Cm: \(\dfrac{x^3}{y^2}+\dfrac{9y^2}{x+2y}\ge4\)

Answer 1

Giải:

Ta có: \(\left(\dfrac{x^3}{y^2}+\dfrac{9y^2}{x+2y}\right)\left(x+x+2y\right)\ge\left(\dfrac{x^2}{y}+3y\right)^2\)

Mặt khác: \(\dfrac{x^2}{y}+3y=\dfrac{2-y^2}{y}+3y=\dfrac{2\left(y^2+1\right)}{y}\ge4\)

Có: \(x+x+2y=2\left(x+y\right)\le2\sqrt{2\left(x^2+y^2\right)}=4\)

\(\Rightarrow\dfrac{x^3}{y^2}+\dfrac{9y^2}{x+2y}\ge\dfrac{\left(\dfrac{x^2}{y}+3y\right)^2}{2x+2y}=\dfrac{4^2}{4}=4\)

Xảy ra khi x = y = 1