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

<=>\((xy+xz+y^2+yz)(x+z)=8xyz\)

<=>\(x^2y+x^2z+y^2z+xyz+xyz+xz^2+z^2y+yz^2=8xyz\)

<=> \(x^2y+x^2z+y^2x+xz^2+y^2z+yz^2-6xyz=0\)

<=> \(y(x^2+z^2-2xz)+x(y^2-2yz+z^2)+z(y^2-2yx+x^2)=0\)

<=>\(y(x-z)^2+x(y-z)^2+z(x-y)^2=0\)

Mà x,y,z dương

=> \((x-z)^2=0=>x=z\)

\((x-y)^2=0=>x=y\)

\((y-z)^2=0=>y=z\)

Vậy x=y=z

Áp dụng BĐT Cauchy cho 2 số không âm:

\(x+y\ge2\sqrt{xy};y+z\ge2\sqrt{yz};x+z\ge2\sqrt{xz}\);

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\sqrt{\left(xyz\right)^2}=8xyz\)

(Dấu "="\(\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\x=z\end{cases}}\Leftrightarrow x=y=z\left(đpcm\right)\))

x+y>=2 căn xy

y+z>=2 căn yz

x+z>=2 căn xz

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

Xét hiệu: (x+y)(y+z)(z+x)-8xyz=0
(=) (x+y)>=2√xy
(y+z)>=2√yz
(z+x)>=2√zx
(=) (x+y)(y+z)(z+x)>=8√x^2 y^2 z^2
(=) (x+y)(y+z)(x+z)>=8|x| |y| |z|
(=) ( x+y)(y+z)(z+x)>= 8xyz

vì x,y,z>0 nên áp dụng bđt côsi ta có

x+y >= 2\(\sqrt{xy}\)

y+z >= 2\(\sqrt{yz}\)

z+x >= 2\(\sqrt{xz}\)

\(\Rightarrow\)(x+y)(y+z)(z+x) >= 8\(\sqrt{x^2y^2z^2}\)

                                >= 8xyz

Dấu = xảy ra <=> x=y=z

Áp dụng BĐT AM - GM ta có: \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}=8xyz\).

Đẳng thức xảy ra khi và chỉ khi x = y = z.

Vậy x = y = z.