Giải thích các bước giải:

$\begin{aligned}
& \dfrac{ab+bc+ca}{abc} = 3 \\
& \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 3 \\
& \text{Đặt } x = \dfrac{1}{a}, y = \dfrac{1}{b}, z = \dfrac{1}{c} \text{ với } x, y, z > 0 \\
& x + y + z = 3 \\
& a+b+c = \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \\
& \dfrac{a}{b} = \dfrac{\dfrac{1}{b}}{\dfrac{1}{a}} = \dfrac{y}{x}, \quad \dfrac{b}{c} = \dfrac{z}{y}, \quad \dfrac{c}{a} = \dfrac{x}{z} \\
& 5 \left( \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \right) \ge \dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z} + 4(x+y+z) \\
& 5 \left( \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \right) - \left( \dfrac{y}{x} + \dfrac{z}{y} + \dfrac{x}{z} \right) - 4(x+y+z) \ge 0 \\
& \left( \dfrac{5}{x} - \dfrac{y}{x} - 4x \right) + \left( \dfrac{5}{y} - \dfrac{z}{y} - 4y \right) + \left( \dfrac{5}{z} - \dfrac{x}{z} - 4z \right) \ge 0 \\
& \dfrac{5 - y - 4x^2}{x} + \dfrac{5 - z - 4y^2}{y} + \dfrac{5 - x - 4z^2}{z} \ge 0 \\
& \text{Thay } 5 - y = 2 + (3 - y) = 2 + x + z \\
& \text{Thay } 5 - z = 2 + (3 - z) = 2 + x + y \\
& \text{Thay } 5 - x = 2 + (3 - x) = 2 + y + z \\
& \dfrac{2 + x + z - 4x^2}{x} + \dfrac{2 + x + y - 4y^2}{y} + \dfrac{2 + y + z - 4z^2}{z} \ge 0 \\
& \left( \dfrac{2}{x} + 1 + \dfrac{z}{x} - 4x \right) + \left( \dfrac{2}{y} + \dfrac{x}{y} + 1 - 4y \right) + \left( \dfrac{2}{z} + \dfrac{y}{z} + 1 - 4z \right) \ge 0 \\
& 2 \left( \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} \right) + \left( \dfrac{z}{x} + \dfrac{x}{y} + \dfrac{y}{z} \right) + 3 - 4(x+y+z) \ge 0 \\
& 2(a+b+c) + \left( \dfrac{a}{c} + \dfrac{c}{b} + \dfrac{b}{a} \right) + 3 - 12 \ge 0 \\
& 2(a+b+c) + \dfrac{a}{c} + \dfrac{c}{b} + \dfrac{b}{a} - 9 \ge 0 \\
& \text{Áp dụng bất đẳng thức trung bình cộng - trung bình nhân:} \\
& \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \ge \dfrac{9}{a+b+c} \\
& 3 \ge \dfrac{9}{a+b+c} \\
& a+b+c \ge 3 \\
& 2(a+b+c) \ge 6 \\
& \dfrac{a}{c} + \dfrac{c}{b} + \dfrac{b}{a} \ge 3 \sqrt[3]{\dfrac{a}{c} \cdot \dfrac{c}{b} \cdot \dfrac{b}{a}} \\
& \dfrac{a}{c} + \dfrac{c}{b} + \dfrac{b}{a} \ge 3 \\
& 2(a+b+c) + \dfrac{a}{c} + \dfrac{c}{b} + \dfrac{b}{a} - 9 \ge 6 + 3 - 9 \\
& 2(a+b+c) + \dfrac{a}{c} + \dfrac{c}{b} + \dfrac{b}{a} - 9 \ge 0 \\
& \text{Kết quả: Bất đẳng thức được chứng minh hoàn toàn, dấu bằng xảy ra khi } a=b=c=1
\end{aligned}$