cho x, y ,z là số dương biết \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\)
CMR \(x+y+2z^2\ge6\)
\(2=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{4}{4z}\ge\dfrac{\left(1+1+2\right)^2}{x+y+4z}\ge\dfrac{16}{x+y+2z^2+2}\\ \Rightarrow x+y+2z^2+2\ge8\\ \Rightarrow x+y+2z^2\ge6\)