Mathematica competitions in Croatia

Transforming the right hand side of the inequality we get $$\begin{array}{rl}{a}^2+b^2+c^2+2(a-b+1)& =a^2+2a+1+b^2-2b+1+c^2\\ & =(a+1{)}^2+(b-1{)}^2+c^2.\end{array}$$ From the A–K inequality it follows $$\sqrt{\frac{(a+1{)}^2+(b-1{)}^2+c^2}{3}}\ge \frac{(a+1)+(b-1)+c}{3}$$ i.e. $$(a+1{)}^2+(b-1{)}^2+c^2\ge \frac{1}{3}(a+b+c{)}^2.$$