Let $c_i=|a_i-b_i|$ and $t=\max\{c_1, c_2, \cdots , c_n\}$. WLOG $c_n=t$. Since $\sum_{i=1}^n a_i = \sum_{i=1}^n b_i$, we get $\sum_{i=1}^{n-1} c_i \ge t$. By Cauchy Inequality, $\sum_{i=1}^{n-1} c_i^2\ge \frac{1}{n-1} (c_1+c_2+\cdots +c_{n-1})^2=\frac{1}{n-1}t^2\implies\sum_{i=1}^{n} c_i^2\ge\frac{n}{n-1}t^2$ $\sum_{i=1}^n \frac{(a_{i+1}+b_{i+1})^2}{n(a_i-b_i)^2+4(n-1)\sum_{j=1}^n a_jb_j} \ge \frac{\sum_{j=1}^n (a_j-b_j)^2+4\sum_{j=1}^n a_jb_j}{nt^2+4(n-1)\sum_{j=1}^n a_jb_j} = \frac{\sum_{j=1}^n c_j^2+4\sum_{j=1}^n a_jb_j}{nt^2+4(n-1)\sum_{j=1}^n a_jb_j} \ge \frac{\frac{n}{n-1}t^2+4\sum_{j=1}^n a_jb_j}{nt^2+4(n-1)\sum_{j=1}^n a_jb_j}=\frac{1}{n-1}$ Equality holds for $n=2$ or $a_i=b_i$.

$\sum_{i=1}^{n-1} c_i^2\ge \frac{1}{n-1} (c_1+c_2+\cdots +c_{n-1})^2=\frac{1}{n-1}t^2\implies\sum_{i=1}^{n} c_i^2\ge\frac{n}{n-1}t^2$

Re1gnover wrote: $\sum_{i=1}^{n-1} c_i^2\ge \frac{1}{n-1} (c_1+c_2+\cdots +c_{n-1})^2=\frac{1}{n-1}t^2\implies\sum_{i=1}^{n} c_i^2\ge\frac{n}{n-1}t^2$ Thanks. Edited.

$\sum_{j=1}^{n} c_j^2\ge\frac{n}{n-1}t^2$ or $\sum_{i=1}^{n} c_i^2\ge\frac{n}{n-1}t^2$ ?

?? What is the difference between first and second one?