Tham khảo:

\(A=\dfrac{7yz}{x}+\dfrac{8zx}{y}+\dfrac{9xy}{z}=\dfrac{3yz}{x}+\dfrac{3zx}{y}+\dfrac{4yz}{x}+\dfrac{4xy}{z}+\dfrac{5zx}{y}+\dfrac{5xy}{z}\)

\(bddt:\) \(AM-GM\Rightarrow\left\{{}\begin{matrix}\dfrac{3yz}{x}+\dfrac{3zx}{y}\ge2\sqrt{\dfrac{9yzxz}{xy}}\ge6z\\\dfrac{4yz}{x}+\dfrac{4xy}{z}\ge2\sqrt{\dfrac{16yzxy}{xz}}\ge8y\\\dfrac{5zx}{y}+\dfrac{5xy}{z}\ge2\sqrt{\dfrac{25zxxy}{yz}}\ge10x\\\end{matrix}\right.\)

\(\Rightarrow A\ge10x+8y+6z\)

Ta có: \(6x^2+8xy+11y^2=2\left(x-y\right)^2+\left(2x+3y\right)^2\ge\left(2x+3y\right)^2\)

Tương tự: \(6y^2+8yz+11z^2\ge\left(2y+3z\right)^2\)

\(6z^2+8zx+11x^2\ge\left(2z+3x\right)^2\)

=> \(P\le\frac{x^2+3xy+y^2}{2x+3y}+\frac{y^2+3yz+z^2}{2y+3z}+\frac{z^2+3zx+x^2}{2z+3x}\)

=> \(4P\le\frac{4x^2+12xy+4y^2}{2x+3y}+\frac{4y^2+12yz+4z^2}{2y+3z}+\frac{4z^2+12zx+4x^2}{2z+3x}\)

\(=\frac{\left(2x+3y\right)^2-5y^2}{2x+3y}+\frac{\left(2y+3z\right)^2-5z^2}{2y+3z}+\frac{\left(2z+3x\right)^2-5x^2}{2z+3x}\)

\(=5\left(x+y+z\right)-5\left(\frac{y^2}{2x+3y}+\frac{z^2}{2y+3z}+\frac{x^2}{2z+3x}\right)\)

\(\le5\left(x+y+z\right)-5.\frac{\left(x+y+z\right)^2}{5\left(x+y+z\right)}=4\left(x+y+z\right)\)

Lại có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\)với mọi x; y; z

=> \(4P\le4.\sqrt{9}=12\)

=> \(P\le3\)

Dấu "=" xảy ra <=> x = y = z = 1

Vậy max P = 3 đạt tại x = y = z = 1.

Lần lượt áp dụng bất đẳng thức cô-si ta có: \(x+y\ge2\sqrt{xy};y+z\ge2\sqrt{yz};z+x\ge2\sqrt{zx}.\)
Suy ra: \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}=8xyz.\)
Dấu bằng xảy ra khi x = y = z.

(x + y)(y + z)(x + z) = 8xyz

⇒ (xy + xz + y² + yz)(x + z) - 8xyz = 0

⇒ x²y + xyz + x²z + xz² + y²x + y²z + xyz + yz² - 8xyz = 0

⇒ x²y - 2xyz + yz² + xy² - 2xyz + xz² + x²z - 2xyz + y²z = 0

⇒ y(x - z)² + x(y - z)² + z(x - y)² = 0

mà x, y, z > 0 (gt)

⇒ \(\left\{{}\begin{matrix}\left(x-z\right)^2=0\\\left(y-z\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x-z=0\\y-z=0\\x-y=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=z\\y=z\\x=y\end{matrix}\right.\)

⇒ x = y = z

* Có BĐT : \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\) với $x,y>0$ ( Chứng minh bằng xét hiệu )

Ta có BĐT : \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\Rightarrow\dfrac{x+y}{x^2+y^2}\le\dfrac{2\left(x+y\right)}{\left(x+y\right)^2}=\dfrac{2}{x+y}\)

Chứng minh tương tự khi đó :

\(P\le\dfrac{2}{x+y}+\dfrac{2}{y+z}+\dfrac{2}{z+x}\)

\(\Rightarrow2P\le\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}=2.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=4032\)

\(\Rightarrow P\le2016\)