Deducing Markov inequality from reverse Markov inequality? Let $x$ be random variable, such that $E(x)=0,E(x^2)=1$  and $P(x^2\geq s^2)\geq\displaystyle\frac{C}{s^t}$, where $C>0,s\geq 1 , t>0$.
Let $m<n$ and $m,n$ are natural numbers very big. Let also $L\geq 1$ .
Consider 
$1-(1-\frac{n}{2}P(x^2\geq Ln))^m$.
Assume (*)    $\frac{n}{2}P(x^2\geq Ln)\leq \frac{2c_0}{m}$, where $0 <c_0 <0.6$;
using inequality $(1-y)^m\leq 1-\displaystyle\frac{my}{2}$, valid for $y\in [0, c_0]$ , with natural $n$, we get
$$1-(1-\frac{n}{2}P(x^2\geq Ln))^m\geq\displaystyle\frac{nm}{4}P(x^2\geq Ln),$$
using assumption $P(x^2\geq s^2)\geq\displaystyle\frac{C}{s^t}$ with $s^2=Ln$ , we get
$$1-(1-\frac{n}{2}P(x^2\geq Ln))^m\geq \displaystyle\frac{nm}{4}P(x^2\geq Ln)\geq \frac{mnC'}{(Ln)^{\frac{t}{2}}}.$$
How to show, that if $t\geq 4$, then (*) holds?

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.
