Question

Cho các số thực dương x,y. Tìm Min P=\(\frac{(x+y)^3}{x^2y}\)

Answer 1

\(P=\dfrac{\left(x+y\right)^3}{x^2y}\\ =\dfrac{x^3+3x^2y+3xy^2+y^3}{x^2y}\\ =\dfrac{x}{y}+3+\dfrac{3y}{x}+\dfrac{y^2}{x^2}\\ =\left(\dfrac{y^2}{x^2}+\dfrac{3y}{x}+\dfrac{x}{y}\right)+3\\ =\left(\dfrac{y^2}{x^2}+\dfrac{y}{2x}+\dfrac{y}{2x}+\dfrac{y}{2x}+\dfrac{y}{2x}+\dfrac{y}{2x}+\dfrac{y}{2x}+\dfrac{x}{8y}+\dfrac{x}{8y}+\dfrac{x}{8y}+\dfrac{x}{8y}+\dfrac{x}{8y}+\dfrac{x}{8y}+\dfrac{x}{8y}+\dfrac{x}{8y}\right)+3\\ \overset{Cauchy}{\ge}15\sqrt[15]{\dfrac{y^2}{x^2}\left(\dfrac{y}{2x}\right)^6\left(\dfrac{x}{8y}\right)^8}+3\\ =\dfrac{15}{4}+3\\ =\dfrac{27}{4}\)

Vậy \(MinP=\dfrac{27}{4}\Leftrightarrow x=2y\).