Đáp án:
$\begin{array}{l}
1)\\
\left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) \ge 2\sqrt {ab} .2\sqrt {bc} .2\sqrt {ac} = 8abc\\
Dấu = xảy\,ra \Leftrightarrow a = b = c\\
2)\\
a)y = \frac{{{x^3} + 1}}{{{x^2}}} = x + \frac{1}{{{x^2}}} = \frac{1}{2}x + \frac{1}{2}x + \frac{1}{{{x^2}}} \ge 3\sqrt[3]{{\frac{1}{2}x.\frac{1}{2}x.\frac{1}{{{x^2}}}}} = \frac{3}{{\sqrt[3]{4}}}\\
Dấu = xảy\,ra \Leftrightarrow \frac{1}{2}x = \frac{1}{{{x^2}}} \Rightarrow {x^3} = 2 \Rightarrow x = \sqrt[3]{2}\\
\Rightarrow GTNN:y = \frac{3}{{\sqrt[3]{4}}}\\
b)y = \frac{{{x^2} + 4x + 4}}{x} = x + 4 + \frac{4}{x}\\
= 4 + x + \frac{4}{x} \ge 4 + 2\sqrt {x.\frac{4}{x}} = 8\\
Dấu = xảy\,ra \Leftrightarrow x = \frac{4}{x} \Rightarrow x = \pm 2\\
\Rightarrow GTNN:y = 8\\
c)y = \frac{{3x}}{2} + \frac{1}{{x + 1}}\\
= \frac{{3\left( {x + 1} \right) - 3}}{2} + \frac{1}{{x + 1}}\\
= \frac{{3\left( {x + 1} \right)}}{2} + \frac{1}{{x + 1}} - \frac{3}{2} \ge 2\sqrt {\frac{3}{2}} - \frac{3}{2} = \frac{{2\sqrt 6 - 3}}{2}\\
\Rightarrow GTNN:y = \frac{{2\sqrt 6 - 3}}{2}
\end{array}$
