Question

Cho x,y,z > 0 , x+y+z=3 Tìm GTNN của \(P=\sqrt{\dfrac{x^3}{y+3}}+\sqrt{\dfrac{y^3}{z+3}}+\sqrt{\dfrac{z^3}{x+3}}\)

Answer 1

\(P=\sqrt{\dfrac{x^3}{y+3}}+\sqrt{\dfrac{y^3}{z+3}}+\sqrt{\dfrac{z^3}{x+3}}\)

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

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

Xét:

\(\left(\sqrt{x\left(y+3\right)}+\sqrt{y\left(z+3\right)}+\sqrt{z\left(x+3\right)}\right)^2\le\left(1^2+1^2+1^2\right)\left(xy+3x+yz+3y+xz+3z\right)\)

\(=3\left(9+xy+yz+xz\right)\)

\(=27+3\left(xy+yz+xz\right)\le27+\left(x+y+z\right)^2=36\)

\(\Rightarrow\sqrt{x\left(y+3\right)}+\sqrt{y\left(z+3\right)}+\sqrt{z\left(x+3\right)}\le6\)
\(P\ge\dfrac{3}{2}\)

\("="\Leftrightarrow x=y=z=1\)