An Inequality:
(1 + 1^{-3})(1 + 2^{-3})(1 + 3^{-3})...(1 + n^{-3}) < 3
Prove the following inequality for all integer n greater than 0:
(1) | (1 + 1^{-3})(1 + 2^{-3})(1 + 3^{-3})...(1 + n^{-3}) < 3 |
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Copyright © 1996-2017 Alexander Bogomolny
Mathematical induction is a reasonable method to apply to proving (1) "for all n". As is the case with another example, there does not appear to be an obvious way to make induction work for (1).
(1) | (1 + 1^{-3})(1 + 2^{-3})(1 + 3^{-3})...(1 + n^{-3}) < 3 |
However, a strengthened inequality
(2) | (1 + 1^{-3})(1 + 2^{-3})(1 + 3^{-3})...(1 + n^{-3}) < 3 - 1/n |
is easily amenable to an inductive argument. Denote the left-hand side of (2) as A(n). The verification of (2) for
For n = k+1, we have
A(k+1) | = A(k)(1 + (k+1)^{-3}) |
< (3 - 1/k)(1 + 1/(k+1)^{3}) | |
= 3 - 1/k + 3/(k+1)^{3} - 1/[k(k+1)^{3}] | |
= 3 - ((k+1)^{3} - 3k + 1)/[k(k+1)^{3}]. |
Our goal of proving (2) for n = k+1 will be achieved provided
The latter simplifies to
and further to
(3) | k^{2} - k + 2 > 0. |
Since P(x) = x^{2} - x + 2 has no real roots and
We thus obtain an example of two problems - one weaker, the other stronger - of which the weaker problem is more difficult to prove than the stronger one. This fits into the observation that more general problem are often easier than their specific instances.
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- Distances to Three Points on a Circle
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- Same Integral, Three Intervals
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- A Refinement of Turkevich's Inequality
- Dan Sitaru's Exercise with Pi and Ln
- Problem 4165 from Crux Mathematicorum
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Copyright © 1996-2017 Alexander Bogomolny
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