$ x_{n} = 2 \cdot n^2$ My solution: Put $ x_{i} = i^2 \cdot \sec^2 {y_{i}}$ Then it changes to: $ 1^2 \cdot \tan {y_{1}} + 2^2 \cdot \tan {y_{2}} + \cdots + n^2 \cdot \tan {y_{n}} = \frac {1}{2} \cdot (1^2 \cdot \sec^2 {y_{1}} + 2^2 \cdot \sec^2 {y_{2}} + \cdots + n^2 \cdot \sec^2 {y_{n}})$ $ \implies 1^2 \cdot \tan {y_{1}} + 2^2 \cdot \tan {y_{2}} + \cdots + n^2 \cdot \tan {y_{n}}$ $ = \frac {1}{2} \cdot (1^2 \cdot (1 + \tan^2 {y_{1}} + 2^2 \cdot (1 + \tan^2 {y_{2}} + \cdots + n^2 \cdot (1 + \tan^2 {y_{n}})$ Now By AM GM inequality for each term we get the result

$ \sum_{k=1}^nk\sqrt{x_k^2-k^2}=\frac{1}{2}\sum_{k=1}^nx_k\iff \sum_{k=1}^n(\sqrt{x_k-k^2}-k)^2=0\iff x_k=2k^2\ \forall k\in [n]$

By AM-GM inequality , $ \frac{(k^2) + (x_k - k^2 )}{2} >= k \sqrt {(x_k - k^2)}$ $ => \frac{x_k }{ 2} >= k \sqrt {(x_k - k^2)}$ Sum this for all values of k we get the given expression , with the equality sign. This means that equality holds for each equation. Equality holds for AM-GM inequality when , the 2 numbers are equal So $ k^2 = x_k - k^2$ $ x_k =2 k^2$