\(C=\dfrac{1}{z}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge\dfrac{1}{z}.\dfrac{4}{x+y}\)(BĐT cauchy)
\(=\dfrac{4}{z\left(x+y\right)}=\dfrac{4}{z\left(3-z\right)}\ge\dfrac{4}{\dfrac{\left(z+3-z\right)^2}{4}}=\dfrac{16}{9}\)(BĐT cauchy)
dấu = xảy ra: \(\left\{{}\begin{matrix}x=y\\z=3-z\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=0,75\\z=1,5\end{matrix}\right.\)
Kl: 1/k=0,5625