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 \(\Leftrightarrow x=y=z;x,y,z>0\)

Ta có : 

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

\(\Leftrightarrow\)\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^3=0^3\)

\(\Leftrightarrow\)\(\left(\frac{1}{x}\right)^3+\left(\frac{1}{y}\right)^3+\left(\frac{1}{z}\right)^3+3\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{z}+\frac{1}{x}\right)=0\)

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

Lại có : 

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

\(\Rightarrow\)\(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=\frac{-1}{z}\\\frac{1}{y}+\frac{1}{z}=\frac{-1}{x}\\\frac{1}{z}+\frac{1}{x}=\frac{-1}{y}\end{cases}}\)

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

\(\Leftrightarrow\)\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\) ( đpcm ) 

Vậy nếu \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\) thì \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

Chúc bạn học tốt ~ 

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

\(\Rightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=\left(-\frac{1}{z}\right)^3\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{x^2y}+\frac{3}{xy^2}=-\frac{1}{z^3}\)

\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{-3}{x^2y}-\frac{3}{xy^2}=\frac{-3}{xy}.\left(\frac{1}{x}+\frac{1}{y}\right)=\frac{-3}{xy}.-\frac{1}{z}=\frac{3}{xyz}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z};\frac{1}{x}+\frac{1}{z}=-\frac{1}{y};\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\)

\(2\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\left(\frac{1}{x^3}+\frac{1}{y^3}\right)+\left(\frac{1}{x^3}+\frac{1}{z^3}\right)+\left(\frac{1}{y^3}+\frac{1}{z^3}\right)\)

\(=\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}\right)+\left(\frac{1}{x}+\frac{1}{z}\right)\left(\frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{y^2}-\frac{1}{yz}+\frac{1}{z^2}\right)\)

\(=-\frac{1}{z}\left(\frac{1}{x^2}-\frac{1}{xy}+\frac{1}{y^2}\right)-\frac{1}{y}\left(\frac{1}{x^2}-\frac{1}{xz}+\frac{1}{z^2}\right)-\frac{1}{x}\left(\frac{1}{y^2}-\frac{1}{yz}+\frac{1}{z^2}\right)\)

\(=-\frac{1}{x^2z}+\frac{1}{xyz}-\frac{1}{y^2z}-\frac{1}{x^2y}+\frac{1}{xyz}-\frac{1}{yz^2}-\frac{1}{xy^2}+\frac{1}{xyz}-\frac{1}{xz^2}\)

\(=\left(-\frac{1}{x^2z}-\frac{1}{x^2y}\right)+\left(-\frac{1}{xy^2}-\frac{1}{y^2z}\right)+\left(-\frac{1}{xz^2}-\frac{1}{yz^2}\right)+\frac{3}{xyz}\)

\(=-\frac{1}{x^2}\left(\frac{1}{z}+\frac{1}{y}\right)-\frac{1}{y^2}\left(\frac{1}{x}+\frac{1}{z}\right)-\frac{1}{z^2}\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{3}{xyz}\)

\(=-\frac{1}{x^2}\cdot-\frac{1}{x}+-\frac{1}{y^2}\cdot-\frac{1}{y}+-\frac{1}{z^2}\cdot-\frac{1}{z}+\frac{3}{xyz}=\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}+\frac{3}{xyz}\)

\(\Rightarrow2\left(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}\right)=\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}+\frac{3}{xyz}\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)(đpcm)

Lời giải:
Đặt $\frac{1}{x}=a; \frac{1}{y}=b; \frac{1}{z}=c$ thì bài toán trở thành:
Cho $a+b+c=0$. Tính $\frac{a^3+b^3+c^3}{abc}$

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Ta có:

$a+b+c=0\Rightarrow a+b=-c$. Khi đó:

$\frac{a^3+b^3+c^3}{abc}=\frac{(a+b)^3-3ab(a+b)+c^3}{abc}$
$=\frac{(-c)^3-3ab(-c)+c^3}{abc}=\frac{-c^3+3abc+c^3}{abc}=\frac{3abc}{abc}=3$

em hiểu rồi ,em cảm ơn

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

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

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