Question

Cho các số thực dương x, y, z thoả mãn: \(x^2+y^2+z^2=3xyz\). Tìm giá trị lớn nhất của: \(P=\dfrac{x^2}{x^4+yz}+\dfrac{y^2}{y^4+xz}+\dfrac{z^2}{z^4+xy}\)

Answer 1

Từ GT ta có: \(3=\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)

Suy ra \(3\le x+y+z\)

Áp dụng AM-GM:

\(VT\le\dfrac{x^2}{2x^2\sqrt{yz}}+\dfrac{y^2}{2y^2\sqrt{xz}}+\dfrac{z^2}{2z^2\sqrt{xy}}=\dfrac{1}{2}\sum\dfrac{1}{\sqrt{xy}}\)

\(=\dfrac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}{2\sqrt{xyz}}\le\dfrac{\sqrt{3\left(x+y+z\right)}}{2\sqrt{xyz}}\le\dfrac{1}{2}\sqrt{\dfrac{\left(x+y+z\right)^2}{xyz}}\)

\(\le\dfrac{1}{2}\sqrt{\dfrac{3\left(x^2+y^2+z^2\right)}{xyz}}=\dfrac{3}{2}\)

Vậy \(P_{Max}=\dfrac{3}{2}\)