Áp dụng bất đẳng thức Cô-si, ta có:

• \[{(a + b)^2} \ge 4ab\]\[ \Leftrightarrow \frac{{a + b}}{4} \ge \frac{{ab}}{{a + b}}\,\,\,\,\,(1)\]

• \[{(b + c)^2} \ge 4bc\]\[ \Leftrightarrow \frac{{b + c}}{4} \ge \frac{{bc}}{{b + c}}\,\,\,\,\,(2)\]

• \[{(c + a)^2} \ge 4ac\]\[ \Leftrightarrow \frac{{c + a}}{4} \ge \frac{{ca}}{{c + a}}\,\,\,\,\,(3)\]

Cộng 3 vế (1); (2) và (3) ta có:

\[\frac{{a + b}}{4} + \frac{{b + c}}{4} + \frac{{c + a}}{4} \ge \frac{{ab}}{{a + b}} + \frac{{bc}}{{b + c}} + \frac{{ca}}{{c + a}}\]

Hay \[\frac{{ab}}{{a + b}} + \frac{{bc}}{{b + c}} + \frac{{ca}}{{c + a}} \le \frac{{(a + b) + (b + c) + (c + a)}}{4}\]

Suy ra \[\frac{{ab}}{{a + b}} + \frac{{bc}}{{b + c}} + \frac{{ca}}{{c + a}} \le \frac{{2(a + b + c)}}{4}\]

Suy ra \[\frac{{ab}}{{a + b}} + \frac{{bc}}{{b + c}} + \frac{{ca}}{{c + a}} \le \frac{{a + b + c}}{2}\]   (đpcm)

Dấu “=” xảy ra khi a = b = c.

Vậy \[\frac{{ab}}{{a + b}} + \frac{{bc}}{{b + c}} + \frac{{ca}}{{c + a}} \le \frac{{a + b + c}}{2}\]