Bài này quen quen
\(x^2+1+y^2+4+z^2+1\ge2x+4y+2z\)
\(\Rightarrow x^2+y^2+z^2\ge2x+4y+2z-6\)
\(\Rightarrow3y\ge2x+4y+2z-6\Leftrightarrow2x+y+2z\le6\)
\(\Rightarrow x+\frac{y}{2}+z\le3\)
\(P=\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(\frac{y}{2}+1\right)^2}+\frac{8}{\left(z+3\right)^2}\)
\(P\ge\frac{1}{2}\left(\frac{1}{x+1}+\frac{1}{\frac{y}{2}+1}\right)^2+\frac{8}{\left(z+3\right)^2}\ge\frac{1}{2}\left(\frac{4}{x+\frac{y}{2}+2}\right)^2+\frac{8}{\left(z+3\right)^2}\)
\(P\ge8\left(\frac{1}{\left(x+\frac{y}{2}+2\right)^2}+\frac{1}{\left(z+3\right)^2}\right)\ge4\left(\frac{1}{x+\frac{y}{2}+2}+\frac{1}{z+3}\right)^2\)
\(P\ge4\left(\frac{4}{x+\frac{y}{2}+z+5}\right)^2\ge4\left(\frac{4}{3+5}\right)^2=1\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;2;1\right)\)
