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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 \(\Leftrightarrow x=y=z;x,y,z>0\)
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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:
$\frac{a^3+b^3+c^3}{abc}=\frac{(a+b)^3-3ab(a+b)+c^3}{abc}$
$=\frac{(-c)^3-3ab(-c)+c^3}{abc}=\frac{-c^3+3abc+c^3}{abc}=\frac{3abc}{abc}=3$
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\left(x+y+z\right)^2=x^2+y^2+z^2\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zy\right)=x^2+y^2+z^2\)
\(\Rightarrow2\left(xy+yz+zx\right)=0\)
\(\Rightarrow xy+yz+zx=0\)
\(\Rightarrow\frac{xy}{xyz}+\frac{yz}{xyz}+\frac{zx}{xyz}=0\)( Chia 2 vế cho xyz )
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
Ta lại có : \(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\left(\frac{1}{x}+\frac{1}{y}\right)^3-\left(\frac{3}{x^2y}+\frac{3}{xy^2}\right)+\frac{1}{z^3}\)
\(=\left(-\frac{1}{z}\right)^3-\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{z^3}\)
\(=-\frac{3}{xy}\cdot-\frac{1}{z}\)\(=\frac{3}{xyz}\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\) ( đpcm )
\(\left(x+y+z\right)^2=x^2+y^2+z^2\)
\(\Leftrightarrow xy+yz+zx=0\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
Ta lại co:
\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}-\frac{3}{xyz}=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}-\frac{1}{xy}-\frac{1}{yz}-\frac{1}{zx}\right)=0\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
TC \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)
\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}\right)^3=-\frac{1}{z^3}\)
\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{3}{xy}\left(\frac{1}{x}+\frac{1}{y}\right)=-\frac{1}{z^3}\)
\(\Leftrightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\left(do\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\right)\)
thay vào \(xyz.\frac{3}{xyz}=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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