Mathematica competitions in Croatia

We enumerate the children by numbers $1,2,\dots ,20$. Let us denote by $d_i$ the number of ribbons that are being held by the child with number $i$. Since each ribbon is being held by two children we have that $$\sum _{i=1}^{20}{d}_i=2\cdot 100=200.$$ The number of all pairs of ribbons is $\frac{100\cdot 99}{2}=4950$. A pair of ribbons such that its four ends are not being held by different children is given by a pair of ribbons such that there is a child who holds one end of each of the two ribbons in that pair. Hence $$\sum _{i=1}^{20}\frac{d_i\cdot (d_i-1)}{2}=4950-4050=900.$$ From here it follows that $$\sum _{i=1}^{20}{d}_i^2=1800+\sum _{i=1}^{20}{d}_i=1800+200=2000.$$ By the AM–QM inequality we have $$10=\frac{d_1+d_2+\cdots +d_{20}}{20}\le \sqrt{\frac{d_1^2+d_2^2+\cdots +d_{20}^2}{20}}=10.$$ Since the equality in AM–QM inequality is obtained if and only if all $d_i$ are equal, the claim follows.