Balkan Mathematical Olympiad Shortlisted Problems

Let us denote
$$L(x_1, x_2, x_3, x_4) = \frac{x_1 + 3x_2}{x_2 + x_3} + \frac{x_2 + 3x_3}{x_3 + x_4} + \frac{x_3 + 3x_4}{x_4 + x_1} + \frac{x_4 + 3x_1}{x_1 + x_2}.$$
Notice that this function is cyclic, i.e. $L(x_1, x_2, x_3, x_4) = L(x_2, x_3, x_4, x_1) = L(x_3, x_4, x_1, x_2) = L(x_4, x_1, x_2, x_3)$. Hence we can suppose that $x_1 \ge x_3$ and $x_4 \ge x_2$. We can also multiply all of the variables with a positive constant without changing its value, i.e. $L(x_1, x_2, x_3, x_4) = L(c \cdot x_1, c \cdot x_2, c \cdot x_3, c \cdot x_4)$.

First we prove that $L(u + v, 0, u, 1) \ge 8$ for positive numbers $u$ and $v$. Indeed
$$L(u + v, 0, u, 1) = \frac{u+v}{u} + \frac{3u}{u+1} + \frac{u+3}{1+u+v} + \frac{1+3u+3v}{u+v} =$$
$$1 + \frac{v}{u} + 3 + \frac{-3}{u+1} + 1 + \frac{2-v}{1+u+v} + 3 + \frac{1}{u+v} =$$
$$8 + \frac{v}{u} - \frac{v}{1+u+v} + \frac{-3}{u+1} + \frac{2}{1+u+v} + \frac{1}{u+v} =$$
$$8 + \frac{v(1+v)}{u(1+u+v)} + \frac{-2v}{(u+1)(1+u+v)} + \frac{1-v}{(u+1)(u+v)} =$$
$$8 + \frac{v(1+v)}{u(u+1)(1+u+v)} + \frac{v(1+v)}{(u+1)(1+u+v)} + \frac{-2v}{(u+1)(1+u+v)} + \frac{1-v}{(u+1)(u+v)} =$$
$$8 + \frac{v(1+v)}{u(u+1)(1+u+v)} + \frac{v(v-1)}{(u+1)(1+u+v)} + \frac{1-v}{(u+1)(u+v)} \ge$$
$$8 + \frac{v(1+v)}{(u+v)(u+1)(1+u+v)} + \frac{(v-1)(v(u+v)-(1+u+v))}{(u+1)(1+u+v)(u+v)} =$$
$$8 + \frac{v(1+v) - (v-1) + (v-1)^2(u+v)}{(u+1)(1+u+v)(u+v)} =$$
$$8 + \frac{v^2 + 1 + (v-1)^2(u+v)}{(u+1)(1+u+v)(u+v)} \ge 8.$$
Also
$$L(u, 0, v, 0) = \frac{u}{v} + \frac{3v}{v} + \frac{v}{u} + \frac{3u}{u} = 3 + \left(\frac{u}{v} + \frac{v}{u}\right) + 3 \ge 6 + 2 \cdot \sqrt{\frac{u}{v} \cdot \frac{v}{u}} = 8.$$
For a constant $c$ we have:
$$L(x_1, x_2, x_3, x_4) - L(x_1 + c, x_2 - c, x_3 + c, x_4 - c) =$$
$$\left( \frac{x_1 + 3x_2}{x_2 + x_3} + \frac{x_2 + 3x_3}{x_3 + x_4} + \frac{x_3 + 3x_4}{x_4 + x_1} + \frac{x_4 + 3x_1}{x_1 + x_2} \right) -$$
$$\left( \frac{x_1 + c + 3(x_2 - c)}{x_2 + x_3} + \frac{x_2 - c + 3(x_3 + c)}{x_3 + x_4} + \frac{x_3 + c + 3(x_4 - c)}{x_4 + x_1} + \frac{x_4 - c + 3(x_1 + c)}{x_1 + x_2} \right) =$$
$$\frac{2c}{x_2 + x_3} - \frac{2c}{x_3 + x_4} + \frac{2c}{x_4 + x_1} - \frac{2c}{x_1 + x_2} =$$
$$2c \left( \frac{x_4 - x_2}{(x_2 + x_3)(x_3 + x_4)} - \frac{x_4 - x_2}{(x_4 + x_1)(x_1 + x_2)} \right) =$$
$$\frac{2c(x_4 - x_2)((x_4 + x_1)(x_1 + x_2) - (x_2 + x_3)(x_3 + x_4))}{(x_2 + x_3)(x_3 + x_4)(x_4 + x_1)(x_1 + x_2)} =$$
$$\frac{2c(x_4 - x_2)(x_1(x_1 + x_2 + x_4) + x_2x_4 - x_3(x_3 + x_2 + x_4) - x_2x_4)}{(x_2 + x_3)(x_3 + x_4)(x_4 + x_1)(x_1 + x_2)} =$$
$$\frac{2c(x_4 - x_2)(x_1 - x_3)(x_1 + x_3 + x_2 + x_4)}{(x_2 + x_3)(x_3 + x_4)(x_4 + x_1)(x_1 + x_2)} =$$
Now if $x_2 = x_4$ and $c = x_2$ we have $L(x_1, x_2, x_3, x_4) = L(x_1 + x_2, 0, x_3 + x_2, 0) \ge 8$.

Similarly for $x_4 > x_2$ and $c = x_2$ we have $L(x_1, x_2, x_3, x_4) \ge L(x_1 + x_2, 0, x_3 + x_2, x_4 - x_2)$. Using $x_1 \ge x_3$ we get
$$\frac{x_1 + x_2}{x_4 - x_2} \ge \frac{x_3 + x_2}{x_4 - x_2}$$
and
$$L(x_1 + x_2, 0, x_3 + x_2, x_4 - x_2) = L\left(\frac{x_1 + x_2}{x_4 - x_2}, 0, \frac{x_3 + x_2}{x_4 - x_2}, 1\right) =$$
$$L(u + v, 0, u, 1) \ge 8,$$
where $u = \frac{x_3+x_2}{x_4-x_2}$ and $v = \frac{x_1-x_3}{x_4-x_2}$.

With this we proved that
$$\frac{x_1 + 3x_2}{x_2 + x_3} + \frac{x_2 + 3x_3}{x_3 + x_4} + \frac{x_3 + 3x_4}{x_4 + x_1} + \frac{x_4 + 3x_1}{x_1 + x_2} \ge 8.$$
for all positive real numbers $x_1, x_2, x_3$ and $x_4$.