An Inequality with Exponents from a Calculus Lemma

Lemma

For $u\gt 0,\;$ $v\in (0,1),\;$ the following inequality holds:

$\displaystyle u^v\gt\frac{u}{u+v}.$

Proofs of the lemma and a list of additional applications have been placed on a separate page.

Proof

Under the conditions of the statement, $a,a+b,a+b+c\in (0,1),\;$ making the application of Lemma possible:

$\displaystyle a^b\gt \frac{a}{a+b},\\ \displaystyle (a+b)^c\gt \frac{a+b}{a+b+c},\\ \displaystyle (a+b+c)^d\gt \frac{a+b+c}{a+b+c+d}.$

The product of the three is exactly the statement to be proved.

Generalization

For $\displaystyle a_1,a_2,\ldots,a_{n-1}\in\left(0,\frac{1}{n-1}\right)\;$ and $a_n\in (0,1),$

$\displaystyle\prod_{k=1}^{n-1}\left(\sum_{i=1}^{k}a_i\right)^{a_{i+1}}\gt\frac{a_1}{\displaystyle\sum_{j=1}^{n}a_j}.$

Acknowledgment

The problem, with the above solution, has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu.