Interesting problem

$\begin{aligned} S & = \sum_{n=1}^{2014} \sqrt{1 + \frac 1{n^2} + \frac 1{(n+1)^2}} \\ & = \sum_{n=1}^{2014} \sqrt{\frac {k^2(k+1)^2+(k+1)^2+k^2}{k^2(k+1)^2}} \\ & = \sum_{n=1}^{2014} \sqrt{\frac {k^4+2k^3+3k^2+2k+1}{k^2(k+1)^2}} \\ & = \sum_{n=1}^{2014} \sqrt{\frac {(k^2+k+1)^2}{k^2(k+1)^2}} \\ & = \sum_{n=1}^{2014} \frac {k^2+k+1}{k(k+1)} \\ & = \sum_{n=1}^{2014} \left(1+\frac 1{k(k+1)}\right) \\ & = \sum_{n=1}^{2014} \left(1+\frac 1k - \frac 1{k+1}\right) \\ & = 2014 + \frac 11 - \frac 1{2015} \\ & = \frac {2015^2-1}{2015} \\ & = \frac {(2015+1)(2015-1)}{2015} \\ & = \frac {2016\cdot2014}{2015} \end{aligned}$

$\implies a+b+c = 2016+2014+2015 = \boxed{6045}$

s can be written as $\sum_{n=1}^{2014} \sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}}$ upon some expanding $\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}}= \sqrt{\dfrac{(n^2+n)^2+(n+1)^2+n^2}{(n^2+n)^2}}=\dfrac{\sqrt{(n^2+n)^2+(n+1)^2+n^2}}{n^2+n}$ upon more expnding $\sqrt{(n^2+n)^2+(n+1)^2+n^2}=\sqrt{n^4+2n^3+3n^2+2n+1}$ what a symmetry. now lets see a chart of some expansions we can see the symmetry for any such expansion. it could easily be proved why. so,we find that $\sqrt{n^4+2n^3+3n^2+2n+1}=n^2+n+1$ now plug this in the original expression to get $\dfrac{\sqrt{(n^2+n)^2+(n+1)^2+n^2}}{n^2+n}=\dfrac{n^2+n+1}{n^2+n}$ expand $\dfrac{n^2+n+1}{n^2+n}=1+\dfrac{1}{n^2+n}=1+\dfrac{1}{n}-\dfrac{1}{n+1}$ take this back to the summation to get $\sum_{n=1}^{2014} (1+\dfrac{1}{n}-\dfrac{1}{n+1})$ write it as $\sum_{n=1}^{2014}(1)+\sum_{n=1}^{2014}(\dfrac{1}{n})-\sum_{n=1}^{2014}(\dfrac{1}{n+1})$ it becomes $2014+(1+\dfrac{1}{2}+\dfrac{1}{3}...... +\dfrac{1}{2013}+\dfrac{1}{2014})-(\dfrac{1}{2}+\dfrac{1}{3}.......+\dfrac{1}{2014}+\dfrac{1}{2015})$ after massive cancellation,it becomes $2014+1-\dfrac{1}{2015}=2015-\dfrac{1}{2015}=\dfrac{2015^2 -1}{2015}$ using the identity $a^2-b^2=(a+b)(a-b)$ $\dfrac{2015^2-1}{2015}=\dfrac{2014\times 2016}{2015}$ and $2014+2016+2015=3\times 2015=6045$