Source: *College Mathematics Journal*, January 1998, p. 70

**Hint:** Don't over-complicate the situation...focus on relationship between x^{2} and x^{a}...

**Solution Commentary:** First, M.J. (Bellingham) notes regarding the first "false" claim: "The key, that I think is missing, is that abs(x^{3}) dominates x^{4} for small x, so 3x^{3}+3x^{4} < 0 for small negative x AND 3x^{3} is smaller than 2x^{3} for neg values.

Counterexample: try x= -.3 to see that (1+x^{2})^{3} > (1+x^{3})^{2}.

Furthermore, eliminating common terms, we are really comparing 2x^{3} with 3x^{3} +4x^{4}. Setting these expressions equal, we can see that they have the same values at x = -1/3 and x=0. So when -1/3 < x < 0, the statement is false."

Second, Beyond M.J.'s claims, the problem can actually be generalized even further...

__Prove:__ For 0 ≤ b < a and 0 < x < 1, (1+x^{a})^{b} ≤ (1+x^{b})^{a}

x^{a} < x^{b}, so (1+x^{a})^{b} < (1+x^{b})^{b} < (1+x^{b})^{a}.

__Note:__ I like this problem because it makes one think carefully about powers of positive numbers, whether less than 1 or greater than 1.