first of all define $ A_i$ as: $ A_i = \frac {a_i^2 + a_{i + 1}^2 - a_{i + 2}^2}{a_i + a_{i + 1} - a_{i + 2}}$ (with $ a_{n + 1} = a_1,a_{n + 2} = a_2$) now we have: $ \begin{eqnarray*}A_i & = & a_i + a_{i + 1} + a_{i + 2} - 2\cdot\frac {a_{i}a_{i + 1}}{a_i + a_{i + 1} - a_{i + 2}} \\ & = & a_i + a_{i + 1} + a_{i + 2} - 2\left(2 + \frac {(a_i - 2)(a_{i + 1} - 2) + 2(a_i + a_{i + 1}) - 4}{a_i + a_{i + 1} - a_{i + 2}}\right)$ now note that $ 1\leq a_i + a_{i + 1} - a_{i + 2}\leq 4$ and $ 0\leq (a_i - 2)(a_{i + 1} - 2)$,thus: $ A_i\leq a_i + a_{i + 1} + a_{i + 2} - 2\left(2 + \frac {2a_{i + 2} - 4}{4}\right) = a_i + a_{i + 1} - 2$ which yields us: $ \sum_{i = 1}^n A_i\leq 2s - 2n$ QED

We can write the $\text{LHS}$ as $\sum_{cyc}\frac{x_i^2+x_{i+1}^2-x_{i+2}^2}{x_i+x_{i+1}-x_{i+2}}$. We can write this as $(x_i + x_{i+1} + x_{i+2}) - \frac{2xixi+1}{x_i + x_{i+1} - x_{i+2}}$. Now $(x_i-2)(x_{i+1}-2)\ge 0$, so $-2x_{i}x_{i+1}\le -4(x_{i}+x_{i+1}-2) = -4((x_i+x_{i+1}-x_{i+2})+(x_{i+2}-2))$. Hence the ith term is less than or equal to $(x_i+x_{i+1}+x_{i+2})-4\left(1+\frac{x_{i+2}-2}{x_i+x_{i+1}-x_{i+2}}\right)$. But the constraints on $x_i$ imply that $(x_i+x_{i+1}-x_{i+2})$ and $(x_{i+2}-2)$ are positive and $(x_i+x_{i+1}-x_{i+2})\le 4$. So the ith term is less than $(x_i+x_{i+1}+x_{i+2})-4\left(1+\frac{x_{i+2}-2}{4}\right)=x_i+x_{i+1}-2$. Hence the $\text{LHS}\le 2s-2n$. We are done. $\Box$