Estonija, 2010

Let the opposite sides of the angles $\alpha$, $\beta$, and $\gamma$ be correspondingly $a$, $b$, and $c$. Since $\cot^2 \alpha = 1/\sin^2 \alpha - 1$ and from the law of sines $1/\sin \alpha = 2R/a$, we have $\cot^2 \alpha = 4R^2/a^2 - 1$; similarly $\cot^2 \beta = 4R^2/b^2 - 1$ and $\cot^2 \gamma = 4R^2/c^2 - 1$. The inequality can therefore be written as
$$4R^2 \cdot \left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right) - 3 \ge 3 \cdot \left( \frac{4 \cdot 9R^2}{(a+b+c)^2} - 1 \right),$$
or
$$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \ge \frac{27}{(a+b+c)^2}.$$
Dividing both sides by 3 and taking the square root gives
$$\sqrt{\frac{1}{3} \cdot \left( \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \right)} \ge \frac{3}{a+b+c}.$$
The left side is the quadratic mean of $\frac{1}{a}$, $\frac{1}{b}$, $\frac{1}{c}$ and the right side is the harmonic mean of the same numbers, hence the inequality holds.
The equality holds iff $a = b = c$.