Croatian Mathematical Olympiad

By the Cauchy-Bunyakovsky-Schwarz inequality, we have $$\frac{(a-1{)}^2}{b}+\frac{(b-1{)}^2}{c}\ge \frac{(2-a-b{)}^2}{b+c}=\frac{c^2}{b+c},$$ and similar inequalities hold for all pairs. By adding the inequalities, we get $$\frac{(a-1{)}^2}{b}+\frac{(b-1{)}^2}{c}+\frac{(c-1{)}^2}{a}\ge \frac{1}{2}\left(\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}\right).$$ Note that the initial claim follows from this, because $$\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a},$$ which holds since $$\begin{array}{rl}\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{c+a}-\frac{a^2}{a+b}-\frac{b^2}{b+c}-\frac{c^2}{c+a}& =\frac{b^2-a^2}{a+b}+\frac{c^2-b^2}{b+c}+\frac{a^2-c^2}{c+a}\\ & =(b-a)+(c-b)+(a-c)=0.\end{array}$$