It's not very hard. We have $(\sum_{k=1}^n a_{k}^{2n})(\sum_{k=1}^n a_{k}^{-2n})-n^2= \sum_{1 \leq i<j \leq n}(\frac{a_{i}^{2n}}{a_{j}^{2n}}+\frac{a_{j}^{2n}}{a_{i}^{2n}})+n-n^2= \sum_{1 \leq i<j \leq n} (\frac{a_{i}^{n}}{a_{j}^{n}}-\frac{a_{j}^{n}}{a_{i}^{n}})^2$. Now it is easy to see that for $x\ge1$, we have $f(x): =x^n-\frac1{x^n}- n(x-\frac1x)\ge0\ge f(\frac1x)$. Indeed, $f^\prime(x)\ge0 \Leftrightarrow (x^n-x)(x^{n+1}-1)\ge0$ for all $x>0$ and $f(1)=0.$ This means for all $x>0$ $(x^n-\frac1{x^n})^2\ge n^2(x-\frac1x)^2$. Put $x=\frac{a_i}{a_j}$ and sum up. Done.

nhat wrote: Let $a_{1},a_{2},...,a_{n}$ be positive real number $(n \geq 2)$,not all equal,such that $\sum_{k=1}^n a_{k}^{-2n}=1$,prove that: $\sum_{k=1}^n a_{k}^{2n}-n^2.\sum_{1 \leq i<j \leq n}(\frac{a_{i}}{a_{j}}-\frac{a_{j}}{a_{i}})^2 >n^2$ For Lagrange's identity, $\sum_i{a_i^{2n}}\sum_j{a_j^{-2n}} - n^2 = \sum_{1 \le i < j \le n}{{\left(\left(\frac{a_i}{a_j}\right)^n - \left(\frac{a_j}{a_i}\right)^n\right)}^2}$ So we only need to prove $\forall x \in \mathbb{R^+}$ $\left(x^n - x^{-n}\right)^2 \ge n^2\left(x - x^{-1}\right)^2$ It's equivalent to $x^{n-1} + x^{n-3} + \cdots + x^{-(n-1)} \ge n$ By AM-GM, we know that $x^r + x^{-r} \ge 2$.