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: $$L(x_1,x_2,x_3,x_4)-L(x_1+c,x_2-c,x_3+c,x_4-c)=$$ $$\left(\frac{x_1+3x_2}{x_2+x_3}+\frac{x_2+3x_3}{x_3+x_4}+\frac{x_3+3x_4}{x_4+x_1}+\frac{x_4+3x_1}{x_1+x_2}\right)-$$ $$\left(\frac{x_1+c+3(x_2-c)}{x_2+x_3}+\frac{x_2-c+3(x_3+c)}{x_3+x_4}+\frac{x_3+c+3(x_4-c)}{x_4+x_1}+\frac{x_4-c+3(x_1+c)}{x_1+x_2}\right)=$$ $$\frac{2c}{x_2+x_3}-\frac{2c}{x_3+x_4}+\frac{2c}{x_4+x_1}-\frac{2c}{x_1+x_2}=$$ $$2c\left(\frac{x_4-x_2}{(x_2+x_3)(x_3+x_4)}-\frac{x_4-x_2}{(x_4+x_1)(x_1+x_2)}\right)=$$ $$\frac{2c(x_4-x_2)((x_4+x_1)(x_1+x_2)-(x_2+x_3)(x_3+x_4))}{(x_2+x_3)(x_3+x_4)(x_4+x_1)(x_1+x_2)}=$$ $$\frac{2c(x_4-x_2)(x_1(x_1+x_2+x_4)+x_2{x}_4-x_3(x_3+x_2+x_4)-x_2{x}_4)}{(x_2+x_3)(x_3+x_4)(x_4+x_1)(x_1+x_2)}=$$ $$\frac{2c(x_4-x_2)(x_1-x_3)(x_1+x_3+x_2+x_4)}{(x_2+x_3)(x_3+x_4)(x_4+x_1)(x_1+x_2)}=$$ 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 $x_4$.