Croatia_2018

Since $x_i\in [0,1]$, we conclude that $x_i\ge x_i^2$, i.e. $$4(x_1+x_2+\cdots +x_n)\ge 4(x_1^2+x_2^2+\cdots +x_n^2).$$ Denote $S=x_1+x_2+\cdots +x_n$. Notice that, because of the previous bound, it suffices to show that $$(S+1{)}^2\ge 4S.$$ This inequality obviously holds since it is equivalent to $(S-1{)}^2\ge 0$.