67th Romanian Mathematical Olympiad

The inequality $$\frac{3}{2}\le \frac{b+c}{b+c+2a}+\frac{a+c}{a+c+2b}+\frac{a+b}{a+b+2c}$$ is Nesbitt's inequality for the triple $(b+c,a+c,a+b)$.

Triangle's inequality yields $b+c>a\iff 3a+3b+3c>4a+2b+2c\iff \frac{2}{3(a+b+c)}<\frac{1}{2a+b+c}\iff \frac{4a}{3(a+b+c)}<\frac{2a}{2a+b+c}\iff \frac{4a}{3(a+b+c)}<1-\frac{b+c}{2a+b+c}$. In the same way $\frac{4b}{3(a+b+c)}<1-\frac{a+c}{2b+a+c}$ and $\frac{4c}{3(a+b+c)}<1-\frac{a+b}{2c+a+b}$. Adding the last three inequalities gives $$\frac{4}{3}<3-\left(\frac{b+c}{b+c+2a}+\frac{a+c}{a+c+2b}+\frac{a+b}{a+b+2c}\right),$$ which is equivalent to what we had to prove.