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Ta có \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xyz}\left(x+y+z\right)=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xyz}=4\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)(vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>0\))
Mặt khác, ta có : \(\frac{1}{x+y+z}=2\) .
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
=> x+y = 0 hoặc y + z = 0 hoặc z + x = 0
Từ đó suy ra P = 0 (lí do vì x,y,z là các số mũ lẻ)
Áp dụng bất đẳng thức Cô-si, ta có: \(\left(3x+1\right)\left(y+z\right)+x=3xy+3xz+\left(x+y+z\right)\ge3xy+3xz+3\sqrt[3]{xyz}\)\(=3xy+3xz+3\Rightarrow\frac{1}{\left(3x+1\right)\left(y+z\right)+x}\le\frac{1}{3\left(xy+xz+1\right)}\)
Tiếp tục áp dụng bất đẳng thức dạng \(u^3+v^3\ge uv\left(u+v\right)\), ta được: \(\frac{1}{3\left(xy+xz+1\right)}=\frac{1}{3\left[x\left(\left(\sqrt[3]{y}\right)^3+\left(\sqrt[3]{z}\right)^3\right)+1\right]}\le\frac{1}{3\left[x\sqrt[3]{yz}\left(\sqrt[3]{y}+\sqrt[3]{z}\right)+1\right]}\)\(=\frac{\sqrt[3]{xyz}}{3\left[\sqrt[3]{x^2}\left(\sqrt[3]{y}+\sqrt[3]{z}\right)+\sqrt[3]{xyz}\right]}=\frac{\sqrt[3]{yz}}{3\left(\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}\right)}\)
Tương tự rồi cộng lại theo vế, ta được: \(P\le\frac{1}{3}\)
Đẳng thức xảy ra khi x = y = z = 1
\(H=\frac{1}{\left(x+1\right)^2+y^2+1}+\frac{1}{\left(y+1\right)^2+z^2+1}+\frac{1}{\left(z+1\right)^2+x^2+1}\)
\(\Leftrightarrow\)\(H=\frac{1}{\left(x+1\right)^2+\left(y+1\right)^2-2y}+\frac{1}{\left(y+1\right)^2+\left(z+1\right)^2-2z}+\frac{1}{\left(z+1\right)^2+\left(x+1\right)^2-2x}\)
Áp dụng BĐT AM-GM ta có:
\(H\le\frac{1}{2.\left(x+1\right)\left(y+1\right)-2y}+\frac{1}{2.\left(y+1\right)\left(z+1\right)-2z}+\frac{1}{2.\left(z+1\right)\left(x+1\right)-2x}\)
\(\Leftrightarrow H\le\frac{1}{2.\left(x+y+xy+1\right)-2y}+\frac{1}{2.\left(y+z+yz+1\right)-2z}+\frac{1}{2.\left(x+z+xz+1\right)-2x}\)
\(\Leftrightarrow H\le\frac{1}{2.\left(x+xy+1\right)}+\frac{1}{2.\left(y+yz+1\right)}+\frac{1}{2.\left(z+xz+1\right)}\)
\(\Leftrightarrow H\le\frac{1}{2}\left[\frac{xyz}{x\left(1+y+yz\right)}+\frac{1}{y+yz+1}+\frac{xyz}{xz\left(y+yz+1\right)}\right]\)
\(\Leftrightarrow H\le\frac{1}{2}\left[\frac{yz}{1+y+yz}+\frac{1}{y+yz+1}+\frac{y}{y+yz+1}\right]=\frac{1}{2}.1=\frac{1}{2}\)
Dấu " = " xảy ra <=> \(x=y=z=1\)
Vậy \(H_{max}=\frac{1}{2}\Leftrightarrow x=y=z=1\)