NMO Selection Tests for the Junior Balkan Mathematical Olympiad

Since $0=(a_1+a_2+a_3+a_4+a_5{)}^2=a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+2{\sum }_{i<j}{a}_i{a}_j$, it follows that ${\sum }_{i<j}(a_i-a_j{)}^2=4(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2)-2{\sum }_{i<j}{a}_i{a}_j=5(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2)$. On the other side, ${\sum }_{i<j}(a_i-a_j{)}^2\le {\sum }_{i<j}|a_i-a_j|$, for $|a_i-a_j|\le 1$. Assuming $a_1\le a_2\le a_3\le a_4\le a_5$, we have ${\sum }_{i<j}|a_i-a_j|=4(a_5-a_1)+2(a_4-a_2)\le 4+2=6$. So $5(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2)\le 6$, as requested to be shown.