Question

Cho \(x,y,z>0\) và \(x+y+z=6\). Tìm max: \(P=\dfrac{xy^2}{y^2+2}+\dfrac{yz^2}{z^2+2}+\dfrac{zx^2}{x^2+2}\)

Answer 1

\(\dfrac{xy^2}{y^2+2}=\dfrac{xy^2}{\dfrac{y^2}{2}+\dfrac{y^2}{2}+2}\le\dfrac{xy^2}{3\sqrt[3]{\dfrac{y^4}{2}}}=\dfrac{1}{3}x\sqrt[3]{2y^2}\le\dfrac{1}{9}x\left(2+y+y\right)=\dfrac{2}{9}\left(x+xy\right)\)

Tương tự: \(\dfrac{yz^2}{z^2+2}\le\dfrac{2}{9}\left(y+yz\right)\) ; \(\dfrac{zx^2}{x^2+2}\le\dfrac{2}{9}\left(z+zx\right)\)

Cộng vế:

\(P\le\dfrac{2}{9}\left(x+y+z+xy+yz+zx\right)\le\dfrac{2}{9}\left(x+y+z+\dfrac{1}{3}\left(x+y+z\right)^2\right)=4\)

Dấu "=" xảy ra khi \(x=y=z=2\)