Question

Cho các số thực dương tm

x² +y²+z²≤3y

Tim min P=1/(x+1)²+4/(y+2)²+8/(z+3)²

Answer 1

\(3y+6\ge x^2+1+y^2+4+z^2+1\ge2x+4y+2z\)

\(\Rightarrow2\left(x+z\right)+y\le6\Rightarrow x+z+\frac{y}{2}\le3\)

\(P=\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(\frac{y}{2}+1\right)^2}+\frac{8}{\left(z+3\right)^2}\ge\frac{1}{2}\left(\frac{1}{x+1}+\frac{1}{\frac{y}{2}+1}\right)^2+\frac{8}{\left(z+3\right)^2}\)

\(P\ge\frac{1}{2}\left(\frac{4}{x+\frac{y}{2}+2}\right)^2+\frac{8}{\left(z+3\right)^2}=8\left(\frac{1}{\left(x+\frac{y}{2}+2\right)^2}+\frac{1}{\left(z+3\right)^2}\right)\)

\(P\ge4\left(\frac{1}{x+\frac{y}{2}+2}+\frac{1}{z+3}\right)^2\ge4\left(\frac{4}{x+\frac{y}{2}+z+5}\right)^2\ge\frac{64}{\left(3+5\right)^2}=1\)

\(P_{min}=1\) khi \(x=\frac{y}{2}=z=1\)