\(P=x+y^2+1+z^3+1+1-3\ge x+2y+3z-3\)
b> ta có
\(x+\frac{1}{x}\ge2\)
\(2y+\frac{2}{y}\ge4\)
\(3z+\frac{3}{z}\ge6\)
Từ đó \(P\ge2+4+6-6-3=2\)
a/ Ta có:
\(P+3=x+\left(y^2+1\right)+\left(z^3+1+1\right)\ge x+2y+3z\)
\(\Rightarrow P\ge x+2y+3z-3\)
b/ Ta có:
\(6=\frac{1}{x}+\frac{2}{y}+\frac{3}{z}=\frac{1}{x}+\frac{4}{2y}+\frac{9}{3z}\)
\(\ge\frac{36}{x+2y+3z}\)
\(\Rightarrow x+2y+3z\ge6\)
\(\Rightarrow P\ge6-3=3\)