Vietnamese Mathematical Competitions

Putting $x = a + b + c$ then we have $a + b + c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 8 \ge \frac{9}{a + b + c} + 8 \Rightarrow x \ge \frac{9}{x} + 8$.
Direct manipulation leads to $(x - 9)(x + 1) \ge 0$ or $x \ge 9$.
Moreover, $a + b + c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 8 < \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + 8 = \frac{19}{2}$.
Thus $9 \le x < \frac{19}{2}$. Now, we rewrite the given condition as:
$$2x = \frac{2}{a} + \frac{2}{b} + \frac{2}{c} + 16 \Leftrightarrow \frac{a - 2}{a} + \frac{b - 2}{b} + \frac{c - 2}{c} = 19 - 2x.$$
By Cauchy-Schwarz inequality, we have
$$\frac{a - 2}{a} + \frac{b - 2}{b} + \frac{c - 2}{c} \ge \frac{[(a - 2) + (b - 2) + (c - 2)]^2}{a(a - 2) + b(b - 2) + c(c - 2)} = \frac{(x - 6)^2}{a^2 + b^2 + c^2 - 2x}.$$
Thus, using previous equation, we have
$$19 - 2x \ge \frac{(x - 6)^2}{a^2 + b^2 + c^2 - 2x} \Rightarrow a^2 + b^2 + c^2 \ge 2x + \frac{(x - 6)^2}{19 - 2x}.$$
In the other hands, the required inequality is equivalent to
$$54 \ge 2(ab + bc + ca) \Leftrightarrow 54 + (a^2 + b^2 + c^2) \ge x^2.$$
By previously proved inequality, it is sufficient to prove that
$$54 + 2x + \frac{(x - 6)^2}{19 - 2x} \ge x^2.$$
Simple calculation leads us to the equivalent inequality
$$(x - 9)(x^2 - 2x - 59) \ge 0,$$
which is true due $x^2 - 2x - 59 = (x - 1)^2 - 60 \ge 8^2 - 60 > 0$.
The equality holds in both sides if and only if $a = b = c = 3$.