\(\begin{array}{l}a)\,\,\,y = \frac{x}{3} + \frac{5}{{2x - 1}};\,\,\,x > \frac{1}{2}\\y = \frac{{2x - 1}}{{2.3}} + \frac{5}{{2x - 1}} + \frac{1}{6}\\\,\,\,\,\, = \frac{{2x - 1}}{6} + \frac{5}{{2x - 1}} + \frac{1}{6}\end{array}\)
Với \(x > \frac{1}{2} \Rightarrow 2x - 1 > 0 \Rightarrow \frac{{2x - 1}}{6};\,\,\frac{5}{{2x - 1}} > 0\)
Áp dụng bất đẳng thức Cô-si cho hai số dương \(\frac{{2x - 1}}{6};\,\,\frac{5}{{2x - 1}}\) ta có:
\(\begin{array}{l}\frac{{2x - 1}}{6} + \frac{5}{{2x - 1}} \ge 2\sqrt {\frac{{2x - 1}}{6}.\frac{5}{{2x - 1}}} = 2\sqrt {\frac{5}{6}} = \frac{{2\sqrt {30} }}{6} = \frac{{\sqrt {30} }}{5}.\\ \Rightarrow y = \frac{{2x - 1}}{6} + \frac{5}{{2x - 1}} + \frac{1}{6} \ge \frac{{\sqrt {30} }}{5} + \frac{1}{6} = \frac{{6\sqrt {30} + 5}}{{30}}\end{array}\)
Dấu ‘‘=’’ xảy ra \( \Leftrightarrow \frac{{2x - 1}}{6} = \frac{5}{{2x - 1}}\)
\(\begin{array}{l} \Leftrightarrow {\left( {2x - 1} \right)^2} = 30 \Leftrightarrow 2x - 1 = \sqrt {30} \\ \Leftrightarrow 2x = \sqrt {30} + 1 \Leftrightarrow x = \frac{{\sqrt {30} + 1}}{2}\,\,\,\left( {tm} \right).\end{array}\)
Vậy \(Min\,\,y = \frac{{\sqrt {30} }}{5}\) khi \(x = \frac{{\sqrt {30} + 1}}{2}.\)
\(\begin{array}{l}b)\,\,\,y = \frac{{{x^2} + 4x + 4}}{x},\,\,\,\,x > 0.\\ \Rightarrow y = x + 4 + \frac{4}{x} = x + \frac{4}{x} + 4.\end{array}\)
Áp dụng bất đẳng thức Cô-si cho hai số dương \(x,\,\,\frac{4}{x}\) ta có:
\(\begin{array}{l}x + \frac{4}{x} \ge 2\sqrt {x.\frac{4}{x}} = 2\sqrt 4 = 4\\ \Rightarrow y = x + \frac{4}{x} + 4 \ge 4 + 4 = 8.\end{array}\)
Dấu ‘‘=’’ xảy ra \( \Leftrightarrow x = \frac{4}{x} \Leftrightarrow {x^2} = 4 \Leftrightarrow x = 2\,\,\,\,\left( {tm} \right).\)
Vậy \(Min\,\,y = 8\,\,\,khi\,\,x = 2.\)
