Giải thích các bước giải:

Ta có : $a\le b\le c\to\dfrac 1a\ge \dfrac 1b\ge \dfrac 1c$ 

$\to 1+\dfrac 1a\ge 1+\dfrac 1b\ge 1+\dfrac 1c$

$\to 2\le (1+\dfrac{1}{a})^3$

$\to 1+\dfrac{1}{a}\ge \sqrt[3]{2}$

$\to 0<a\le 3$

$\to a\in\{1,2,3\}$

$+)a=1\to (1+\dfrac 1b)(1+\dfrac 1c)=1$

$\to 1\le (1+\dfrac 1b)^2$

Mà $b>0\to (1+\dfrac 1b)^2>1\to$ vô nghiệm

$+)a=2\to (1+\dfrac 1b)(1+\dfrac 1c)=\dfrac 43$

$\to\dfrac{4}{3}\le (1+\dfrac{1}{b})^2\to 0<b\le 6$

$\to b\in\{1,2,3,4,5,6\}\to c\in\{3,9,loại,-15,-9,-7\}$

$\to (a,b,c)\in\{(2,1,3),(2,2,9)\}$

$+)a=3\to (1+\dfrac 1b)(1+\dfrac 1c)=\dfrac 32$

$\to \dfrac{3}{2}\le (1+\dfrac{1}{b})^2\to 1\le b\le 4$

$\to b\in\{1,2,3,4\}\to c\in\{4,loại,-8,-5\}$

$\to (a,b,c)=(3,1,4)$