Question

cho x,y,z>0 và x+y+z<=3.Tìm gtnn của P=x^2+y^2+z^2+20/x+y+z

Answer 1

\(P=x^2+y^2+z^2+\dfrac{20}{x+y+z}\ge\dfrac{\left(x+y+z\right)^2}{3}+\dfrac{20}{x+y+z}\)

\(\Leftrightarrow P\ge\dfrac{\left(x+y+z\right)^2}{3}+\dfrac{9}{x+y+z}+\dfrac{9}{x+y+z}+\dfrac{2}{x+y+z}\)

\(\Leftrightarrow P\ge3\sqrt[3]{\dfrac{\left(x+y+z\right)^2}{3}.\dfrac{9}{x+y+z}.\dfrac{9}{x+y+z}}+\dfrac{2}{3}\)

 (theo AM-GM và do \(x+y+z\le3\Rightarrow\dfrac{2}{x+y+z}\ge\dfrac{2}{3}\))

\(\Leftrightarrow P\ge\dfrac{29}{3}\)

Dấu = xảy ra khi x=y=z=1

Vậy minP\(=\dfrac{29}{3}\)