Giải thích các bước giải:
Ta có:
$M = \dfrac{a}{{a + b + c}} + \dfrac{b}{{a + b + d}} + \dfrac{c}{{b + c + d}} + \dfrac{d}{{a + c + d}}$
Do $a,b,c,d\in N*$ nên ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{a}{{a + b + c}} < \dfrac{a}{{a + b}}\\
\dfrac{b}{{a + b + d}} < \dfrac{b}{{a + b}}
\end{array} \right.\\
\Rightarrow \dfrac{a}{{a + b + c}} + \dfrac{b}{{a + b + d}} < \dfrac{a}{{a + b}} + \dfrac{b}{{a + b}} = \dfrac{{a + b}}{{a + b}} = 1
\end{array}$
Tương tự:
$\begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{c}{{b + c + d}} < \dfrac{c}{{c + d}}\\
\dfrac{d}{{a + c + d}} < \dfrac{d}{{c + d}}
\end{array} \right.\\
\Rightarrow \dfrac{c}{{b + c + d}} + \dfrac{d}{{a + c + d}} < \dfrac{c}{{c + d}} + \dfrac{d}{{c + d}} = \dfrac{{c + d}}{{c + d}} = 1
\end{array}$
Như vậy:
$\begin{array}{l}
\dfrac{a}{{a + b + c}} + \dfrac{b}{{a + b + d}} + \dfrac{c}{{b + c + d}} + \dfrac{d}{{a + c + d}} < 2\\
\Rightarrow M < 2\\
\Rightarrow {M^{10}} < {2^{10}} < 2021
\end{array}$
