The sum of five positive numbers is $100$. Prove that there are two numbers among them whose difference is at most $10$.
attempt: Let $m,n,p,q,r$ be positive real numbers such that $m \leq n \leq p \leq q \leq r$. Then assume $m > 0, n > 10, p> 20, q > 30 , e > 40.$ Can anyone please suggest something ? Thank you