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?
 A: 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.
A: Let us first recall Markov inequality: let $Z$ denote a nonnegative integrable random variable. Then, for every positive $z$, 
$$
\mathrm P(Z\geqslant z)\leqslant z^{-1}\mathrm E(Z).
$$
Application: consider $Z=X^2$ for a square integrable random variable $X$ such that $\mathrm E(X^2)=1$, and $z=n$ for any positive integer $n$. Then $\mathrm P(X^2\geqslant n)\leqslant n^{-1}$.
Note that this proof uses neither the hypothesis that $\mathrm E(X)=0$ nor any lower bound on the tail of the distribution of $X^2$.

Now, what you seem to be asking for is a proof that $\mathrm P(X^2\geqslant n)\leqslant\Theta(n^{-2})$ under the additional hypotheses that $\mathrm E(X)=0$ and that $\mathrm P(X^2\geqslant s)\geqslant Cs^{-t/2}$ for every $s\geqslant1$, for some $t\geqslant4$.
But there is no hope such a result could hold, is there? Consider a symmetric random variable such that $\mathrm P(X^2\geqslant s)=\Theta(s^{-u})$ when $s\to\infty$, for some positive $u$. Then your hypothesis may hold as soon as $u\leqslant t/2$ (for the tail estimate) and $u\gt1$ (for the square integrability) and your conclusion asks that $u\geqslant2$. These simply do not fit.
