Let's assume that Praneeth is wrong, and try to find four numbers that work.

1)  Assume the four distinct numbers a b c d are in geometric progression.  Then a b d  and a c d are not in geometric progression.  Since the arithmetic mean of a and d cannot be both b and c, it follows that at least one of these is neither an arithmetic or geometric progression.

Therefore, we cannot satisfy the problem requirements by having all 4 in geometric progression. (Similarly, we cannot satisfy the problem requirements by having all 4 in arithmetic progression.)

    

Therefore, there must be at least one arithmetic progression among the 4 numbers.

    

2)  Assume the arithmetic progression is  e f g.  Call the fourth number x.  As shown above, it cannot be in an arithmetic series with e f g.

    

If x < e, then then order is x e f g.  Because at most one of e and f is the geometric mean between e and x, at least one of x e g and x f g is neither arithmetic nor geometric.

    

    Similarly, if the order is e x f g,  at least one of e x f or e x g is not geometric.

    

    If the order is e f x g, then at least one of e x g or f x g is not geometric.

    

    And if the order is e f g x, then at least one of e f x or e g x is not geometric.

    

 3) Therefore, it is not possible to find four numbers that work, and the Praneeth's assertion is proved.