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

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

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

1) (12x^2-12xy+3y^2)-10x(2x-y)+8=3(2x-y)^2-10x(2x-y)+8=(2x-y)(6x-3y-10x)+8=8-(2x-3y)(4x+3y)

2) áp dụng BĐT cauchy ta có (x+y)(y+z)(z+x)\(\ge\)\(2\sqrt{xy}\).\(2\sqrt{yz}\).\(2\sqrt{xz}\)=8xyz

dấu đẳng thức xảy ra khi x=y=z

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