\(F=x+y+\frac{1}{2x}+\frac{2}{y}\)
\(F=\frac{x}{2}+\frac{x}{2}+\frac{y}{2}+\frac{y}{2}+\frac{1}{2x}+\frac{2}{y}\)
\(F=\left(\frac{x}{2}+\frac{1}{2x}\right)+\left(\frac{x}{2}+\frac{y}{2}\right)+\left(\frac{y}{2}+\frac{2}{y}\right)\)
\(F=\frac{1}{2}\left(x+\frac{1}{x}\right)+\left(\frac{x+y}{2}\right)+\left(\frac{y}{2}+\frac{2}{y}\right)\)
Ta có: \(x+\frac{1}{x}\ge2\Rightarrow\frac{1}{2}\left(x+\frac{1}{x}\right)\ge1\left(1\right)\)
\(x+y\ge3\Rightarrow\frac{x+y}{2}\ge\frac{3}{2}\left(2\right)\)
\(\frac{y}{2}+\frac{2}{y}\ge2\left(3\right)\)
Cộng lần lượt từng vế của 3 BĐT \(\left(1\right);\left(2\right);\left(3\right)\) ta được:
\(\frac{1}{2}\left(x+\frac{1}{x}\right)+\left(\frac{x+y}{2}\right)+\left(\frac{y}{2}+\frac{2}{y}\right)\ge1+\frac{3}{2}+2=\frac{9}{2}\)
\(\Rightarrow F\ge\frac{9}{2}\)
Vậy \(Min_F=\frac{9}{2}\)
\(F=\frac{x}{2}+\frac{1}{2x}+\frac{y}{2}+\frac{2}{y}+\frac{1}{2}\left(x+y\right)\)
\(F\ge2\sqrt{\frac{x}{4x}}+2\sqrt{\frac{2y}{2y}}+\frac{1}{2}.3=\frac{9}{2}\)
\(\Rightarrow F_{min}=\frac{9}{2}\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
