## 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) $\displaystyle (1 + 1^{-3})(1 + 2^{-3})(1 + 3^{-3})\ldots (1 + n^{-3}) \lt 3$

Solution

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) $\displaystyle (1 + 1^{-3})(1 + 2^{-3})(1 + 3^{-3})\ldots (1 + n^{-3}) \lt 3$

However, a strengthened inequality

 (2) $(1 + 1^{-3})(1 + 2^{-3})(1 + 3^{-3})\ldots (1 + n^{-3}) \lt 3 - 1/n$

is easily amenable to an inductive argument. Denote the left-hand side of (2) as $A(n).$ The verification of (2) for $n = 1$ is immediate. Assume (2) holds for $n = k:$

$A(k) \lt 3 - 1/k$

For n = k+1, we have

\displaystyle \begin{align} A(k+1)&= A(k)(1 + (k+1)^{-3})\\ &\lt (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}].\\ \end{align}

Our goal of proving (2) for $n = k+1$ will be achieved provided

$((k+1)^{3} - 3k + 1)/[k(k+1)^{3}] \gt 1/(k+1).$

The latter simplifies to

$k^{3} + 3k^{2} + 2 \gt k(k + 1)^{2},$

and further to

 (3) $k^{2} - k + 2 \gt 0.$

Since $P(x) = x^{2} - x + 2$ has no real roots and $P(0) = 2 \gt 0,$ $P(x) \gt 0,$ for all real $x,$ i.e., (3) holds and, with it, (2) for $n = k+1.$ The induction is complete.

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.

### References

1. T. Andreescu, B. Enescu, Mathematical Olympiad Treasures, Birkhäuser, 2004