A Nice Algebra Problem (if such exists)

Theorem

If positive reals $x$ and $y$ are such that their sum $s=x+y$ is a constant, then their product $p=xy$ increases when their absolute difference $|x-y|$ decreases.

The denominators of the terms in the summation is of the form $k(2014-k)$ . If we let $x=k$ and $y=2014-k$ , then the condition $x+y=s$ or $k + 2014-k = 2014$ , where $s$ is a constant, is met. Therefore, we can use the theorem (see proof below) above. We note that the absolute difference $|2k-2014|$ of the denominator decreases from the two ends (first and last) inward. That is $\{|2k-2014|\} =$ $\{2012, 2010, 2008, ... ,2012\}$ . This means that all the denominators within is greater than the first and last denominators which is 2013. This means that all the terms within is smaller the first and last terms which is $\frac 1{2013}$ .

$\begin{aligned} S & = \frac 1{1\cdot 2013} + \frac 1{2\cdot 2012} + \frac 1{3\cdot 2011} +\cdots + \frac 1{2013\cdot 1} \\ & \ {\color{#D61F06}<} \ \underbrace {\frac 1{2013} + \frac 1{2013} + \frac 1{2013} + \cdots + \frac 1{2013}}_{\text{2013 terms}} = \end{aligned}$

Yes, the sum is smaller than 1.

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Proof

Let $x+y=s$ , where $s$ is a constant, and $p=xy$ . Also let $d=x-y$ , then $x = \dfrac {s+d}2$ and $x = \dfrac {s-d}2$ and $p = xy = \dfrac {s+d}2 \times \dfrac {s-d}2 = \dfrac {s^2-d^2}4$ . As $s$ is constant, $p$ increases as $d^2$ or $|x-y|$ decreases. $\square$