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Theo giả thiết ta có : \(x+yz=yz-z-1=\left(z-1\right)\left(y+1\right)=\left(x+y\right)\left(y+1\right)\)
Tương tự : \(y+zx=\left(x+y\right)\left(x+1\right)\)
Và \(z+xy=\left(x+1\right)\left(y+1\right)\)
Nên \(P=\frac{x}{\left(x+y\right)\left(y+1\right)}+\frac{y}{\left(x+y\right)\left(x+1\right)}+\frac{z^2+2}{\left(x+1\right)\left(y+1\right)}\)
\(=\frac{x^2+y^2+x+y}{\left(x+y\right)\left(x+1\right)\left(y+1\right)}+\frac{z^2+2}{\left(x+1\right)\left(y+1\right)}\)
Ta có \(x^2+y^2\ge\frac{\left(x+y\right)^2}{2};\left(x+1\right)\left(y+1\right)\le\frac{\left(x+y+2\right)^2}{4}\)
nên \(P\ge\frac{2\left(x+y\right)^2+4\left(x+y\right)}{\left(x+y+2\right)^2\left(x+y\right)}+\frac{4\left(z^2+2\right)}{\left(x+y+2\right)^2}=\frac{2\left(x+y\right)+4}{\left(x+y+2\right)^2}+\frac{4\left(z^2+2\right)}{\left(x+y+2\right)^2}\)
\(=\frac{2}{z+1}+\frac{4\left(z^2+2\right)}{\left(z+1\right)^2}=f\left(z\right);z>1\)
Lập bảng biến thiên ta được \(f\left(z\right)\ge\frac{13}{4}\) hay min \(P=\frac{13}{4}\) khi \(\begin{cases}z=3\\x=y=1\end{cases}\)
Áp dụng bất đẳng thức Cauchy
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ge\frac{9}{xy+yz+zx}\)
\(M\ge\frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+zx}=\frac{1}{x^2+y^2+z^2}+\frac{4}{2\left(xy+yz+xz\right)}+\frac{7}{xy+yz+zx}\)
Áp dụng BĐT Cauchy - Schwarz :
\(\frac{1}{x^2+y^2+z^2}+\frac{4}{2\left(xy+yz+zx\right)}\ge\frac{\left(1+2\right)^2}{\left(x+y+z\right)^2}=9\)
và \(\frac{7}{xy+yz+xz}\ge\frac{7}{\frac{1}{3}\left(x+y+z\right)^2}=21\)
\(\Rightarrow M\ge9+21=30\)
Dấu " = " xảy ra khi \(x=y=z=\frac{1}{3}\)
Áp dụng BĐT Cauchy schwarz ta có:
\(M=\frac{1}{x^2+y^2+z^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)
\(\ge\frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+zx}\)
\(=\frac{1}{x^2+y^2+z^2}+\frac{4}{2\left(xy+yz+zx\right)}+\frac{7}{2\left(xy+yz+zx\right)}\)
\(\ge\frac{9}{\left(x+y+z\right)^2}+\frac{7}{\frac{2\left(x+y+z\right)^2}{3}}=30\)
Đẳng thức xảy ra tại x=y=z=1/3
Áp dụng Bất Đẳng Thức Cosi ta có \(\hept{\begin{cases}\frac{x^3}{1+y}+\frac{1+y}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{x^3}{1+y}\cdot\frac{1+y}{4}\cdot\frac{1}{2}}=\frac{3x}{2}\\\frac{y^3}{1+z}+\frac{1+z}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{y^3}{1+z}\cdot\frac{1+z}{4}\cdot\frac{1}{2}}=\frac{3y}{2}\\\frac{z^3}{1+x}+\frac{1+x}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{z^3}{1+x}\cdot\frac{1+x}{4}\cdot\frac{1}{2}}=\frac{3z}{2}\end{cases}}\)
Cộng vế theo vế ta được \(P+\frac{3+x+y+z}{4}+\frac{3}{2}\ge\frac{3}{2}\left(x+y+z\right)\)
\(\Leftrightarrow P\ge\frac{5}{4}\left(x+y+z\right)-\frac{9}{4}\)
Mà ta có \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\ge9\Rightarrow x+y+z\ge3\)
Do đó \(P\ge\frac{5}{4}\cdot3-\frac{9}{4}=\frac{3}{2}\). Dấu "=" xảy ra khi x=y=z=1
Vậy minP=\(\frac{3}{2}\)khi x=y=z=1
Áp dụng BĐT Cô - si cho 3 bộ số không âm
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)
\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)
\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)
Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)
Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)
Ta có bất đẳng thức: với \(x,y>0\)
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
Dấu \(=\)khi \(x=y\).
Áp dụng bất đẳng thức trên ta được:
\(\frac{1}{2x+3y+3z}\le\frac{1}{4}\left(\frac{1}{2x+y+z}+\frac{1}{2y+2z}\right)\le\frac{1}{4}\left[\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)+\frac{1}{2}\left(\frac{1}{y+z}\right)\right]\)
\(=\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)+\frac{1}{8}\left(\frac{1}{y+z}\right)\)
Tương tự với \(\frac{1}{3x+2y+3z},\frac{1}{3x+3y+2z}\)sau đó cộng lại vế với vế ta được:
\(P\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=3\)
Dấu \(=\)xảy ra khi \(x=y=z=\frac{1}{8}\)
đặt \(\left(a;b;c\right)=\left(\sqrt{\frac{yz}{x}};\sqrt{\frac{zx}{y}};\sqrt{\frac{xy}{z}}\right)\)\(\Rightarrow\)\(a^2+b^2+c^2=1\)
\(A=\Sigma\frac{1}{1-ab}=\Sigma\frac{2ab}{2\left(a^2+b^2+c^2\right)-2ab}+3\le\frac{1}{2}\Sigma\frac{\left(a+b\right)^2}{b^2+c^2+c^2+a^2}\)
\(\le\frac{1}{2}\Sigma\left(\frac{a^2}{c^2+a^2}+\frac{b^2}{b^2+c^2}\right)=\frac{9}{2}\)
dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\frac{1}{3}\)