Using contradiction, prove that for every four positive real numbers c, d, e and f, at least one of c, d, e, f is
greater than or equal to the average of c, d, e, f.
I don't believe that there are any relevant equations for this problem. I do know that we have to suppose that the negation of the statement is true, show that this supposition leads to a contradiction and then conclude that the statement to be proved is true.
The Attempt at a Solution
I am finding it difficult to translate the above statement into predicate logic because I have only worked with two variables and there are four in this problem! Here is my attempt but I think that I'm way off.
[itex]\forall[/itex] (c [itex]\wedge[/itex] d [itex]\wedge[/itex] e [itex]\wedge[/itex] f) [itex]\in[/itex] ℝ+, [itex]\exists[/itex](c [itex]\vee[/itex] d [itex]\vee[/itex] e [itex]\vee[/itex] f) [itex]\in[/itex] ℝ+ [itex]\geq[/itex] (c+d+e+f)/4
The negation in words might be something like: at least one of the four positive real numbers c, d, e and f will be less than the average of c, d, e, f.