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Sửa đề : CMR : \(xyz\le\frac{1}{8}\)
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge2\Rightarrow\frac{1}{z+1}\ge\left(1-\frac{1}{x+1}\right)+\left(1-\frac{1}{z+1}\right)\)
\(=\frac{x}{x+1}+\frac{y}{y+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\left(1\right)\)(bđt AM - GM)
Tương tự ta cũng có : \(\hept{\begin{cases}\frac{1}{x+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(y+1\right)}}\left(2\right)\\\frac{1}{y+1}\ge2\sqrt{\frac{xz}{\left(x+1\right)\left(z+1\right)}}\left(3\right)\end{cases}}\)
Nhân vế với vế của (1) ; (2) ; (3) laih ta được :
\(\frac{1}{x+1}.\frac{1}{y+1}.\frac{1}{z+1}\ge8\sqrt{\frac{\left(xyz\right)^2}{\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2}}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\frac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(\Rightarrow xyz\le\frac{1}{8}\)(đpcm)
Xét: \(x+y+z=xyz\Leftrightarrow\frac{x+y+z}{xyz}=1\)
\(\Leftrightarrow\frac{x}{xyz}+\frac{y}{xyz}+\frac{z}{xyz}=1\Leftrightarrow\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}=1\)
Mặt khác:\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\sqrt{3}\)<=>\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\left(\sqrt{3}\right)^2\)
<=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}=3\)
<=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)=3\)
<=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2.1=3\)
<=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2=3\)
<=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1\)
\(\hept{\begin{cases}xyz=12\\x^3+y^3+z^3=36\end{cases}}\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)-3xyz+z^3=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow x=y=z\left(x+y+z>0\right)\)
Thay x=y=z vào r tính thôi bạn
Vì xy + yz + zx = 1 ta có :
\(\frac{x-y}{z^2+1}+\frac{y-z}{x^2+1}+\frac{z-x}{y^2+1}=\frac{x-y}{z^2+xy+yz+zx}+\frac{y-z}{x^2+xy+yz+zx}+\frac{z-x}{y^2+xy+yz+zx}\)
\(=\frac{x-y}{\left(y+z\right)\left(z+x\right)}+\frac{y-z}{\left(x+y\right)\left(x+z\right)}+\frac{z-x}{\left(y+z\right)\left(x+y\right)}\)
\(=\frac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(x+z\right)\left(z-x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{0}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(ĐPCM)