By Cauchy-Schwarz in Engel Form, we have \begin{align*} \sum^{n-1}_{i=1} \frac{1}{P_i + P_{i+1}}&>\frac{(n-1)^2}{2(P_1+P_2+\ldots+P_n)-P_1-P_n} \\&=\frac{(n-1)^2}{2(1+2+\ldots+n)-P_1-P_n} \\&=\frac{(n-1)^2}{n(n+1)-P_1-P_n} \\&\ge\frac{(n-1)^2}{n(n+1)-(n-1)-n} \\&=\frac{(n-1)^2}{n^2-n+1} \\&>\frac{(n-1)^2}{n^2+n-2} \\&=\frac{n-1}{n+2}.\end{align*}

So cute First, the statement is altogether wrong; there are not $n$ permutations $P_1,P_2,\ldots,P_n$, each being an arbitrary permutation of $\{1,2,\ldots,n\}$, since $\dfrac {1} {P_i + P_{i+1}}$ makes no sense as a number, but rather $n$ numbers $P_1,P_2,\ldots,P_n$, being an arbitrary permutation of $\{1,2,\ldots,n\}$. Second, the proof is quite nice, up to a patently reversed inequality; the right form is \begin{align*} \sum^{n-1}_{i=1} \frac{1}{P_i + P_{i+1}}&>\frac{(n-1)^2}{2(P_1+P_2+\cdots+P_n)-P_1-P_n} \\&=\frac{(n-1)^2}{2(1+2+\cdots+n)-P_1-P_n} \\&=\frac{(n-1)^2}{n(n+1)-P_1-P_n} \\&\ge\frac{(n-1)^2}{n(n+1)-1-2} \\&=\frac{(n-1)^2}{n^2+n-3} \\&>\frac{(n-1)^2}{n^2+n-2} \\&=\frac{n-1}{n+2}.\end{align*} Most likely the best such permutation is $1,n,2,n-1,\ldots, \lfloor n/2\rfloor +1$.

Could you use a smoothing argument to turn all the fractions into 1/(n+1)? I turned $p_i < p_j < p_k < p_l $ into $ \frac{1}{p_i+p_j}+\frac{1}{p_k+p_l} > \frac{1}{p_i+p_l} + \frac{1}{p_j+p_k}$ but I'm not sure if one can treat this as an invariant and solve the problem. *Also Cauchy-Schwarz in Engel-Form = Titu's Lemma for those who read the solutions above and were a bit confused on what that was. (#me)

You can see also here for a very similar problem. http://artofproblemsolving.com/community/c6h17331p118699

$\noindent \textit {Sketch}$: use the HM-AM inequality, in which your terms are $P_i+P_{i+1}$, for $i\in \{1, 2\ldots n-1\}$ (a total of $n-1$ terms).