Probably it is better reformulate the question I've posted. Please tell me if my solution is correct and where I have mistakes. Thank you for advice.
Let x be random variable with mean zero and variance 1. And such that for $C>0$, $t>0$ and natural n, $P(x^2\ge n)\geq \frac{C}{n^t}$. We want to show that if $t\ge 2$, then $P(x^2\ge n)\geq \frac{2}{n^2}$.
First, we show that for any $n\geq \widetilde{n}$, $ \begin{align} P(x^2\geq n)\leq \frac{2}{n^2}. \end{align} $ Suppose to the contrary that \begin{align} P(x^2 \geq n_i) > \frac{2}{n_i^2} \end{align} for some infinite sequence $n_1$, $n_2$, $\ldots$, $n_i$, $\ldots$, such that \begin{align} P(x^2 > n_i^2) > 2P(x^2 > n_{i+1}^2). \end{align} By assumption $ 1=E x^2 = \int_{0}^\infty x^2 d \mu. $ Here $\mu$ is a probability measure. We can break up this integral into the sum: $ E x^2 = \sum_{i=0}^\infty \int_{n_i}^{n_{i+1}}x^2 d \mu. $ Consider now $ \begin{align} \int_{n_i}^{n_{i+1}} x^2 d\mu \geq n_i^2 \int_{n_i}^{n_{i+1}} d\mu &= n_i^2(P(x^2\geq n_i^2)-P(x^2 \geq n_{i+1}^2)) &\gt n_i^2\frac 12 P(x^2\geq n_i^2) &\gt 1. \end{align} $ This is a contradiction to $E x^2=1$.
Thus, with $\widetilde{n}\leq n$ \begin{align} P(w^2\geq Kn)\leq \frac {2}{n^2} . \end{align} But, within condition \begin{align} \frac{C}{n^t}\leq P(w^2\geq Kn)\leq \frac{2}{n^2}, \end{align} and $\displaystyle{n\geq \left(\frac{C}{2}\right)^{\frac{1}{t-2}}=\widetilde{\widetilde{n}}}$. Thus, for any $n\geq \max\{\widetilde{n},\widetilde{\widetilde{n}}\}$, $\frac{C}{n^t}\leq P(w^2\geq Kn)\leq \frac{2}{n^2}$.
If $t\geq 2$, then for any $n \in N$ and ,$C=2$ $P(w^2\geq Kn)=\frac{2}{n^2}$. Thus, $\frac{C}{n^t}\leq P(w^2\geq Kn)\leq \frac{2}{n^2}$ for any natural n.