THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD

First solution. Swapping, if necessary, the roles of $a$ and $c$ and those of $b$ and $d$, we may assume that $a\le c$. In this case, the function $f(x)=\frac{a+x}{c+x}$ is increasing on $[1,2]$, while $g(x)=\frac{c+x}{a+x}$ is decreasing on $[1,2]$, hence $\frac{a+b}{b+c}+\frac{c+d}{d+a}\le \frac{a+2}{c+2}+\frac{c+1}{a+1}$. On the other hand, $4\cdot \frac{a+c}{b+d}\ge a+c$, therefore it is sufficient to prove that $\frac{a+2}{c+2}+\frac{c+1}{a+1}\le a+c$, which reduces to $a+c+4\le a^{2}c+a{c}^2+a^2+3ac$. From $(a-1)(c-1)\ge 0$ and $a^{2}c+a{c}^2+a^2+2ac\ge 5$ the conclusion follows, with equality if $a=c=1$, and $b=d=2$.

Second solution. We show that $$\frac{a+b}{b+c}\le 2\cdot \frac{a+c}{b+2}\le 2\cdot \frac{a+c}{b+d}\text{and}\frac{c+d}{d+a}\le 2\cdot \frac{a+c}{b+d},$$ inequalities which give the conclusion. The second inequality can be obtained from the first one by swapping $a$ with $c$ and $b$ with $d$, therefore it is sufficient to prove the first one. This inequality reduces to $b^2+ab+2b+2a\le 2c^2+2ac+2bc+2ab$. But $2a\le 2ac$, $2b\le 2bc$ and $b^2\le b+2\le ab+2c^2$, the inequality $b^2\le b+2$ being equivalent to $(b+1)(b-2)\le 0$. We obtain the equality case $a=1,b=2,c=1$ and $d=2$.

Third solution. As $0<b+d\le 4$, it is sufficient to prove that the left hand side is at most $a+c$. Without loss of generality, we may assume that $a\le c$. In this case, $$\frac{a+b}{b+c}+\frac{c+d}{d+a}\le 1+\left(1+\frac{c-a}{d+a}\right)\le 2+c-a\le c+a,$$ with equality if $a=1,b=2,c=1$ and $d=2$.