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Ta có
\(\hept{\begin{cases}\left(x+1\right)^2\ge0\\\left(y+1\right)^2\ge0\\\left(z+1\right)^2\ge0\end{cases}}\)và \(\hept{\begin{cases}x^2+1>0\\y^2+1>0\\z^2+1>0\end{cases}}\)
\(\Rightarrow A=\frac{\left(x+1\right)^2\left(y+1\right)^2}{z^2+1}+\frac{\left(y+1\right)^2\left(z+1\right)^2}{x^2+1}+\frac{\left(z+1\right)^2\left(x+1\right)^2}{y^2+1}\ge0\)
Kết hợp với điều kiện ban đầu thì
GTNN của A là 0 đạt được khi
\(\left(x,y,z\right)=\left(-1,-1,5;-1,5,-1;5,-1-1\right)\)
We have:
\(A=\Sigma_{cyc}\frac{1}{3xy+3zx+x+y+z}\le\frac{1}{3xy+3zx+3\sqrt[3]{xyz}}=\Sigma_{cyc}\frac{1}{3xy+3zx+3}=\Sigma_{cyc}\frac{1}{3\left(xy+zx+1\right)}\)
Dat \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow abc=1\)
\(\Rightarrow A\le\Sigma_{cyc}\frac{1}{3\left(\frac{1}{ab}+\frac{1}{ca}+1\right)}=\Sigma_{cyc}\frac{a}{3\left(a+b+c\right)}=\frac{1}{3}\)
Dau '=' xay ra khi \(x=y=z=1\)
Từ giả thiết \(x+y+z=xyz\Leftrightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Khi đó \(\frac{x}{1+x^2}=\frac{\frac{1}{x}}{\frac{1}{x^2}+1}=\frac{\frac{1}{x}}{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}=\frac{xyz}{\left(x+y\right)\left(x+z\right)}\)
Tương tự cho 2 cái còn lại ta có: \(\frac{y}{1+y^2}=\frac{xyz}{\left(y+x\right)\left(y+z\right)}\)
\(\frac{z}{1+z^2}=\frac{xyz}{\left(z+x\right)\left(z+y\right)}\)
Suy ra \(VT=\frac{xyz\left(y+z\right)+2xyz\left(z+x\right)+3xyz\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Đpcm
\(2P-2=2\left(xy+yz+zx\right)-2\left(x^2+y^2+z^2\right)+x^2\left(y-z\right)^2+y^2\left(z-x\right)^2+z^2\left(x-y\right)^2\)
\(=-\left(x-y\right)^2-\left(y-z\right)^2-\left(z-x\right)^2+x^2\left(y-z\right)^2+y^2\left(z-x\right)^2+z^2\left(x-y\right)^2\)
\(=\left(x-y\right)^2\left(z^2-1\right)+\left(y-z\right)^2\left(x^2-1\right)+\left(z-x\right)^2\left(y^2-1\right)\le0\)
\(\text{( Do }x^2;y^2;z^2\le1\text{)}\)
\(\Rightarrow2P\le2\Rightarrow P\le1\)
\(\text{Dấu bằng xảy ra khi và chỉ khi 1 trong 3 số bằng 1; 2 số còn lại bằng 0.}\)
Ta có:
\(\frac{x}{1+x^2}+\frac{18y}{1+y^2}+\frac{4z}{1+z^2}=xyz\left(\frac{1}{yz\left(1+x^2\right)}+\frac{18}{xz\left(1+y^2\right)}+\frac{4}{xy\left(1+z^2\right)}\right)\)
\(=xyz\left(\frac{1}{yz+x\left(x+y+z\right)}+\frac{18}{xz+y\left(x+y+z\right)}+\frac{4}{xy+z\left(x+y+z\right)}\right)\)
\(=xyz\left(\frac{1}{\left(x+y\right).\left(x+z\right)}+\frac{18}{\left(y+x\right).\left(y+z\right)}+\frac{4}{\left(z+x\right).\left(z+y\right)}\right)\)
\(=xyz.\frac{\left(z+y\right)+18.\left(x+z\right)+4\left(x+y\right)}{\left(x+y\right).\left(y+z\right).\left(z+x\right)}\)
\(=\frac{xyz\left(22x+5y+19z\right)}{\left(x+y\right).\left(y+z\right).\left(z+x\right)}\)(đpcm)