Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

=>\(\frac{1}{z}=-\left(\frac{1}{x}+\frac{1}{y}\right)\)

=>\(\left(\frac{1}{z}\right)^3=\left[-\left(\frac{1}{x}+\frac{1}{y}\right)\right]^3\)

=>\(\frac{1}{z^3}=-\left[\left(\frac{1}{x}+\frac{1}{y}\right)^3\right]\)

=>\(\frac{1}{z^3}=-\left[\left(\frac{1}{x}\right)^3+3.\left(\frac{1}{x}\right)^2.\frac{1}{y}+3.\frac{1}{x}.\left(\frac{1}{y}\right)^2+\left(\frac{1}{y}\right)^3\right]\)

=>\(\frac{1}{z^3}=-\left[\frac{1}{x^3}+3.\frac{1}{x}.\frac{1}{y}.\frac{1}{x}+3.\frac{1}{x}.\frac{1}{y}.\frac{1}{y}+\frac{1}{y^3}\right]\)

=>\(\frac{1}{z^3}=-\left[\frac{1}{x^3}+3.\frac{1}{x}.\frac{1}{y}.\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{y^3}\right]\)

=>\(\frac{1}{z^3}=-\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\left\{-\left[3.\frac{1}{x}.\frac{1}{y}.\left(\frac{1}{x}+\frac{1}{y}\right)\right]\right\}\)

\(\frac{1}{z^3}-\left[-\left(\frac{1}{x^3}+\frac{1}{y^3}\right)\right]=-\left[3.\frac{1}{x}.\frac{1}{y}.\left(\frac{1}{x}+\frac{1}{y}\right)\right]\)

Vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0=>\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)

=>\(\frac{1}{z^3}+\frac{1}{x^3}+\frac{1}{y^3}=-3.\frac{1}{x}.\frac{1}{y}.\left(-\frac{1}{z}\right)\)

=>\(\frac{1}{z^3}+\frac{1}{x^3}+\frac{1}{y^3}=3.\frac{1}{x}.\frac{1}{y}.\frac{1}{z}\)

=>\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=3.\frac{1}{xyz}\)

=>\(xyz.\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=3\)

=>A=3

Vậy A=3

Với x ; y > 0 , cần c/m : \(x^3+y^3\ge xy\left(x+y\right)\)

Ta có : \(x^3+y^3-xy\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2-xy\right)=\left(x+y\right)\left(x-y\right)^2\ge0\)

( điều này luôn đúng với mọi x ; y > 0 )

=> BĐT được c/m

Áp dụng vào bài toán , ta có :

\(\frac{1}{x^3+y^3+xyz}+\frac{1}{y^3+z^3+xyz}+\frac{1}{x^3+z^3+xyz}\le\frac{1}{xy\left(x+y\right)+xyz}+\frac{1}{yz\left(y+z\right)+xyz}+\frac{1}{xz\left(x+z\right)+xyz}=\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{xz\left(x+y+z\right)}=\frac{x+y+z}{xyz\left(x+y+z\right)}=\frac{1}{xyz}\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=z;x,y,z>0\)

Xét (1/x+1/y+1/z)^2=1/x^2+1/y^2+1/z^2+2/xy+2/yz+2/xz

=P+2/xy+2/yz+2/xz=P+(2z+2x+2y)/xyz=P+2(x+y+z)/x+y+z=P+2

mà (1/x+1/y+1/z)^2=3

=>p=3-2=1

ta có :  a+b+c=0 => (a+b+c)(a²+b²+c²-ab-ac-bc)=0  (đoạn này bạn tự nhân ra rồi rút gọn nhé)

=>  a³+b³+c³-3abc=0  =>  a³+b³+c³= 3abc  

thay a=\(\frac{1}{x}\);b=\(\frac{1}{y}\);c=\(\frac{1}{z}\)

=>\(\frac{1}{x^3}\)+\(\frac{1}{y^3}\)+\(\frac{1}{z^3}\)=3.\(\frac{1}{xyz}\)

A=xyz(\(\frac{1}{x^3}\)+\(\frac{1}{y^3}\)+\(\frac{1}{z^3}\))  =  xyz .3 . \(\frac{1}{xyz}\)=3

\(x^3+y^3+1\ge xy\left(x+y\right)+xyz=xy\left(x+y+z\right)\)

=> \(\frac{1}{x^3+y^3+1}\le\frac{1}{xy\left(x+y+z\right)}\)

Hai cái còn lại tương tự

=>  A \(\le\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{xz\left(x+y+z\right)}=\frac{1}{x+y+z}\cdot\frac{x+y+z}{xyz}=1\)

Vậy MAx A = 1 tại x = y = z = 1 

Lời giải:
a. Xét hiệu:

$x^3+y^3-xy(x+y)=(x^3-x^2y)-(xy^2-y^3)=x^2(x-y)-y^2(x-y)$

$=(x-y)(x^2-y^2)=(x-y)^2(x+y)\geq 0$ với mọi $x,y\geq 0$

$\Rightarrow x^3+y^3\geq xy(x+y)$

Dấu "=" xảy ra khi $x=y$

b.

Áp dụng BĐT phần a vô:

$x^3+y^3\geq xy(x+y)$

$\Rightarrow x^3+y^3+1\geq xy(x+y)+1=xy(x+y)+xyz=xy(x+y+z)$

$\Rightarrow \frac{1}{x^3+y^3+1}\leq \frac{1}{xy(x+y+z)}=\frac{xyz}{xy(x+y+z)}=\frac{z}{x+y+z}$

Hoàn toàn tương tự với các phân thức còn lại suy ra:

$\text{VT}\geq \frac{z}{x+y+z}+\frac{x}{x+y+z}+\frac{y}{x+y+z}=1$

Ta có đpcm

Dấu "=" xảy ra khi $x=y=z=1$