Đáp án:
$MinD = 4 + \sqrt 2 \Leftrightarrow a = 1$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
D = \dfrac{a}{{{a^2} + 1}} + \dfrac{{5\left( {{a^2} + 1} \right)}}{{2a}}\\
= \dfrac{a}{{{a^2} + 1}} + \dfrac{{{a^2} + 1}}{{2a}} + \dfrac{{4\left( {{a^2} + 1} \right)}}{{2a}}\\
= \left( {\dfrac{a}{{{a^2} + 1}} + \dfrac{{{a^2} + 1}}{{2a}}} \right) + \dfrac{{2\left( {{a^2} + 1} \right)}}{a}\\
\ge 2\sqrt {\dfrac{a}{{{a^2} + 1}}.\dfrac{{{a^2} + 1}}{{2a}}} + 2.\dfrac{{2\sqrt {{a^2}.1} }}{a}\left( {bdtCauchy} \right)\\
= \dfrac{2}{{\sqrt 2 }} + \dfrac{{4a}}{a}\left( {a > 0} \right)\\
= 4 + \sqrt 2
\end{array}$
$ \Rightarrow MinD = 4 + \sqrt 2 \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{a}{{{a^2} + 1}} = \dfrac{{{a^2} + 1}}{{2a}}\\
{a^2} = 1
\end{array} \right. \Leftrightarrow {a^2} = 1 \Leftrightarrow a = 1$
Vậy $MinD = 4 + \sqrt 2 \Leftrightarrow a = 1$
