Junior Balkan Mathematical Olympiad

Solution:

The inequality can be written as:
$$\frac{5 + 2(1 + b)}{1 + a} + \frac{5 + 2(1 + c)}{1 + b} + \frac{5 + 2(1 + a)}{1 + c} \geq \frac{69}{4}.$$
We substitute $1 + a = x$, $1 + b = y$, $1 + c = z$.
So, we have to prove the inequality
$$\frac{5 + 2y}{x} + \frac{5 + 2z}{y} + \frac{5 + 2x}{z} \geq \frac{69}{4} \Leftrightarrow 5\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) + 2\left(\frac{y}{x} + \frac{z}{y} + \frac{x}{z}\right) \geq \frac{69}{4}$$
where $x, y, z > 1$ real numbers and $x + y + z = 4$.
We have
- $\frac{x + y + z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \Leftrightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x + y + z} \Leftrightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq \frac{9}{4}$
- $\frac{y}{x} + \frac{z}{y} + \frac{x}{z} \geq 3 \cdot \sqrt[3]{\frac{y}{x} \cdot \frac{z}{y} \cdot \frac{x}{z}} = 3$

Thus,
$$\frac{5 + 2y}{x} + \frac{5 + 2z}{y} + \frac{5 + 2x}{z} = 5\left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) + 2\left(\frac{y}{x} + \frac{z}{y} + \frac{x}{z}\right) \geq 5 \cdot \frac{9}{4} + 2 \cdot 3 = \frac{69}{4}.$$
The equality holds when $(x = y = z, \frac{y}{x} = \frac{z}{y} = \frac{x}{z}, x + y + z = 4)$, thus $x = y = z = \frac{4}{3}$, i.e. $a = b = c = \frac{1}{3}$.