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## Attention! General Eyes Consider...
(1+x3)2 = 1+2x3+x6
(1+x2)3 = 1+3x3+3x4+x6

Thus, we oberve that (1+x3)2 ≤ (1+x2)3
where x can be any real number.

Prove: For 2 ≤ a and 0 < x < 1, (1+xa)2 ≤ (1+x2)a

Note, the restriction 0 < x < 1. It seems necessary now, but wasn't even mentioned in the first example. Should it have been? That is, is the first statement (1+x3)2 ≤ (1+x2)3
"true" (or provable) for any real number x?

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

Hint: Don't over-complicate the situation...focus on relationship between x2 and xa...

Solution Commentary: First, M.J. (Bellingham) notes regarding the first "false" claim: "The key, that I think is missing, is that abs(x3) dominates x4 for small x, so 3x3+3x4 < 0 for small negative x AND 3x3 is smaller than 2x3 for neg values.

Counterexample: try x= -.3 to see that (1+x2)3 > (1+x3)2.

Furthermore, eliminating common terms, we are really comparing 2x3 with 3x3 +4x4. 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+xa)b ≤ (1+xb)a

xa < xb, so (1+xa)b < (1+xb)b < (1+xb)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.