If x = 2 then 2 y < y 2 for y = 3 ; 2 y = y 2 for y = 4 ; 2 y > y 2 for y > 4.
For an induction proof without calculus it looks like the thing to prove is F ( n) > F ( n + 1) for n ≥ 3 , which is the same as n n + 1 > ( n + 1) n, or n > ( 1 + 1 n) n. It is known that ( 1 + 1 n) n increases toward e which is less than 3, which settles the problem. Without relying on that one can replace n − 1 of the factors in ( 1 + 1 / n) n by a telescoping product of terms ( n + 1 − i) / ( n − i), for i = 0 to ( n − 2), which leaves an inequality similar to n > ( n / 2) that can be checked easily and implies the one on F ( n).
As in the real-valued problem usually solved with calculus, the key is the function F ( t) = t 1 / t. The inequalities x y > y x and F ( x) > F ( y) are equivalent, reducing the problem to a comparison of values of F at different points. For positive integers larger than 1, the maximum value is F ( 3), and F () is decreasing starting at 3, but F ( 2) = F ( 4). This implies that the exceptions observed above are the only ones to the pattern that it is more efficient to increase the exponent than the base.
For positive integers, the nontrivial case is when x ≠ y and both are at least 2.
Actually it's easy for x and y positive. Assume that we've labeled x and y so that y > x. Compute t = y/x. Then
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.