If $ a_1,a_2,\cdots,a_n(n\geq2)$ be natural numbers such that $ a_1 < a_2 < \cdots < a_n$ and $ \sum_{k = 1}^{n}\frac {1}{a_k}\leq1$, then the following inequlity $ \left(\sum_{k = 1}^{n}\frac {1}{a_{k}^2 + x}\right)^2\leq \frac {m}{a_1(a_1 - 1) + x}$ holds for all nonnegative real number $ x$ if $ m = 0.3229383\cdots$, which is unique real root of the following irreducible polynomial over $ \mathbb{Q}:$ $ 5180741959680000m^5 - 10225012469299200m^4 + 7842687249241912m^3$ $ - 2829413217657726m^2 + 654489379537533m - 87403180032000;$ with equality if and only if $ n = 3,a_1 = 2,a_2 = 3, a_3 = 6, x = 1.6248\cdots$, which is unique real root of the following irreducible polynomial over $ \mathbb{Q}:$ $ 3x^5 + 159x^4 + 3570x^3 + 24464x^2 + 39456x - 145152.$

consider two cases $a_1(a_1-1)\le{x^2} $ and ${x^2}\le{a_1(a_1-1)}$ the first case is easy , use AM-GM first then use cauchy ,the second part is more difficult,using cauchy first\[\left(\sum_{i=1}^n\frac{1}{a_i^2+x^2}\right)^2\le\frac{1}{2}\cdot\frac{1}{a_1(a_1-1)+x^2} \]$\left(\sum_{i=1}^n\frac{1}{a_i^2+x^2}\right)^2\le$$(\sum_{i=1}^n\frac{1}{a_i})$$(\sum_{i=1}^n\frac{a_i}{(a_i^2+x^2)^2})$ the rest part is easy