Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
a + 1 \ge 2\sqrt a \\
b + 1 \ge 2\sqrt b \\
c + 1 \ge 2\sqrt c
\end{array} \right.\left( {BDT:Cauchy} \right)\\
\Rightarrow \left( {a + 1} \right)\left( {b + 1} \right)\left( {c + 1} \right) \ge 8\sqrt {abc} = 8
\end{array}$
Dấu bằng xảy ra
$ \Leftrightarrow \left\{ \begin{array}{l}
a = 1\\
b = 1\\
c = 1
\end{array} \right.$
Ta có điều phải chứng minh.
b) Ta có:
$\begin{array}{l}
\dfrac{{x + 3}}{{2020}} + \dfrac{{4x + 8}}{{2019}} + \dfrac{{5x + 5}}{{2018}} = 0\\
\Leftrightarrow \dfrac{{x + 3}}{{2020}} + \dfrac{{2\left( {2x + 4} \right)}}{{2019}} + \dfrac{{5\left( {x + 1} \right)}}{{2018}} = 0\\
\Leftrightarrow \dfrac{{x + 3}}{{2020}} + \dfrac{{2\left( {x + 1 + x + 3} \right)}}{{2019}} + \dfrac{{5\left( {x + 1} \right)}}{{2018}} = 0\\
\Leftrightarrow \dfrac{{x + 3}}{{2020}} + \dfrac{{2\left( {x + 3} \right)}}{{2019}} + \dfrac{{2\left( {x + 1} \right)}}{{2019}} + \dfrac{{5\left( {x + 1} \right)}}{{2018}} = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {\dfrac{1}{{2020}} + \dfrac{2}{{2019}}} \right) + \left( {x + 1} \right)\left( {\dfrac{2}{{2019}} + \dfrac{5}{{2018}}} \right) = 0\\
\Leftrightarrow \dfrac{{\left( {x + 3} \right).6059}}{{2020.2019}} + \dfrac{{\left( {x + 1} \right).14131}}{{2019.2018}} = 0\\
\Leftrightarrow \dfrac{{\left( {x + 3} \right).6059}}{{2020}} + \dfrac{{\left( {x + 1} \right).14131}}{{2018}} = 0\\
\Leftrightarrow x.\left( {\dfrac{{6059}}{{2020}} + \dfrac{{14131}}{{2018}}} \right) = - \dfrac{{3.6059}}{{2020}} - \dfrac{{14131}}{{2018}}\\
\Leftrightarrow x = 5,33
\end{array}$
Vậy phương trình có nghiệm duy nhất $x = 5,33$
