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
Best Answer
Assume all numbers have a difference that is bigger than $10$. Let these $5$ numbers be $a>b>c>d>e>0$. So we have $$a>b+10>c+20>d+30>e+40>40$$So, $$a>e+40, \quad b>e+30, \quad c>e+20, \quad d>e+10$$ So $$100=a+b+c+d+e>5e+100>100$$ Contradiction.