Iranian Mathematical Olympiad

First solution. Just to simplify the calculations, define $S = a + c$ and $P = ac$. In this case, the RHS of the inequality becomes
$$RHS = \frac{a+b+c}{3} - \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} = \frac{S+b}{3} - \frac{3Pb}{Sb+P}.$$
The goal is to minimize the above phrase as a function of $b$; Note that the phrase, as a function of $b$, has a linear term which is $\frac{S+b}{3}$, and a fractional term with the denominator $Sb + P$. Now we re-write the phrase using the new variable $t = Sb + P$ and with a little calculation we shall have
$$RHS = \frac{S+b}{3} - \frac{3Pb}{Sb+P} = \frac{S^2 - 10P}{3S} + \frac{t}{3S} + \frac{P^2}{St}.$$
Note that the multiply of the last two terms is a phrase without the variable $t$. Therefore with using the AM-GM inequality we obtain
$$RHS \ge \frac{S^2 - 10P}{3S} + 2\sqrt{\frac{P^2}{S^2}} = \frac{S^2 - 4P}{3S} = \frac{(c-a)^2}{3(c+a)} \ge \frac{(c-a)^2}{6c}. \blacksquare$$

Second solution. For the easement of calculations, for any arbitrary formula $F(a, b, c)$ with three variables $a$, $b$ and $c$, the cyclic sum
$$F(a, b, c) + F(b, c, a) + F(c, a, b),$$
is shown with $\sum F(a, b, c)$. Using this notation, we expand the RHS of the inequality
$$\begin{aligned} \frac{\sum a}{3} - \frac{3}{\sum \frac{1}{a}} &= \frac{\sum a}{3} - \frac{3abc}{\sum ab} = \frac{(\sum a)(\sum ab) - 9abc}{3 \sum ab} \\ &= \frac{\sum (a^2b + ab^2) - 6abc}{3 \sum ab} = \frac{\sum a(b-c)^2}{3 \sum ab}. \end{aligned}$$
Therefore, the claim of the problem is equivalent to the following inequality
$$\sum a(b-c)^2 \ge \left( \frac{\sum ab}{2c} \right) (c-a)^2. \quad (1)$$
But note that with condition $a \le b \le c$, $\frac{ab}{2c}$ is not greater than $\frac{b}{2}$ so for the RHS of the above inequality we have
$$\left( \frac{\sum ab}{2c} \right) (c-a)^2 = \left( \frac{ab}{2c} + \frac{a+b}{2} \right) (c-a)^2 \le \left( \frac{a}{2} + b \right) (c-a)^2. \quad (2)$$
For the LHS of (1), we have
$$\sum a(b-c)^2 \ge b(c-a)^2 + a((c-b)^2 + (b-a)^2). \quad (3)$$
Note that for any two numbers $X, Y$,
$$\begin{aligned} (X - Y)^2 \ge 0 &\implies X^2 + Y^2 - 2XY \ge 0 \\ &\implies 2(X^2 + Y^2) \ge (X + Y)^2 \\ &\implies X^2 + Y^2 \ge \frac{(X + Y)^2}{2}. \end{aligned}$$
Therefore, if $X = c - b$, $Y = b - a$ using (3), we obtain
$$\sum a(b-c)^2 \ge b(c-a)^2 + a((c-b)^2 + (b-a)^2) \ge b(c-a)^2 + \frac{a}{2}(c-a)^2.$$
This inequality along with (2) implies (1) which is equivalent to the claim of the problem. ■