Đáp án:
\(\sqrt {2018} + \sqrt {2020} < 2\sqrt {2019} \)
Giải thích các bước giải:
\(\begin{array}{l}
{\left( {\sqrt {2018} + \sqrt {2020} } \right)^2} = 2018 + 2020 + 2\sqrt {2018.2020} \\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4038 + 2\sqrt {2018.2020} \\
{\left( {2\sqrt {2019} } \right)^2} = 4.2019 = 4038 + 2.\sqrt {{{2019}^2}} \\
Ta\,\,so\,\,sanh\,\,2018.2020\,\,va\,\,{2019^2}\\
2018.2020 = \left( {2019 - 1} \right)\left( {2019 + 1} \right) = {2019^2} - 1 < {2019^2}\\
\Rightarrow 2018.2020 < {2019^2}\\
\Rightarrow \sqrt {2018.2020} < \sqrt {{{2019}^2}} \\
\Rightarrow 2\sqrt {2018.2020} < 2\sqrt {{{2019}^2}} \\
\Rightarrow 4038 + 2\sqrt {2018.2020} < 4038 + 2.\sqrt {{{2019}^2}} \\
\Rightarrow {\left( {\sqrt {2018} + \sqrt {2020} } \right)^2} < {\left( {2\sqrt {2019} } \right)^2}\\
\Rightarrow \sqrt {2018} + \sqrt {2020} < 2\sqrt {2019}
\end{array}\)
