If A is the average of three positive real numbers then one of the three numbers is less than or equal to A.

This'll get it teed up for you: To prove by contradiction, first assume the premise is true; i.e...

• A is the average of three positive reals s, t, and u.

But then also assume the negation of the consequent is true...

• s > A, t > A, u > A

Then show how the second assumption renders the first assumption impossible. Hint: add the three inequalities together.