Hellenic Mathematical Olympiad

For the set $X$, we use the symbol $|X|$ for the number of its elements. The number of all possible differences $\alpha -\beta$, with $(\alpha ,\beta )\in \Gamma$, is at most equal to the number of the elements of the set $\Gamma$, that is $$d\le |\Gamma |.(\ast )$$ Let $A=\{a_1,a_2,...,a_m\}$. For each $i$, we suppose that there exist $b_i$ pairs $(a_i,b)\in \Gamma$. Then there exist $b_i^2$ triads $(a_i,\lambda ,\mu )$, with $(a_i,\lambda )\in \Gamma$ and $(a_i,\mu )\in \Gamma$. Therefore, there exist totally $p=b_1^2+\cdots +b_m^2$ triads $(\kappa ,\lambda ,\mu )$ with $(\kappa ,\lambda )\in \Gamma ,(\kappa ,\mu )\in \Gamma$. From Cauchy-Schwarz inequality we get $$m(b_1^2+\cdots +b_m^2)\ge (b_1+\cdots +b_m{)}^2(1)$$ The left hand side is equal to $|A|\cdot p$, while the right hand side is equal to $|\Gamma {|}^2$. Since it is given that $|A|\le n$, by using relations (1) and (*) we get: $$n\cdot p\ge |\Gamma {|}^2\ge d^2.$$