Mathematica competitions in Croatia

$$\frac{a^2}{a+b}+\frac{a+b}{4}\ge a\text{and}\frac{b^2}{b+c}+\frac{b+c}{4}\ge b.$$ Adding these two inequalities we get $$\left(\frac{a^2}{a+b}+\frac{a+b}{4}\right)+\left(\frac{b^2}{b+c}+\frac{b+c}{4}\right)\ge a+b=\frac{3a+2b-c}{4}+\frac{a+b}{4}+\frac{b+c}{4}.$$ If we subtract $\frac{a+b}{4}+\frac{b+c}{4}$ from both sides, the result follows.