is it by Chebychev's inequality? Let x be random variables with mean zero and variance 1.
Let n be natural number, t>0, C>0.
Let also $P(x^2\geq n)\geq C/(n^t)$.
Show, if $t\geq4$, then $P(x^2\geq n)\leq C/(n^2)$.

I got so fare, that with $t\geq 4$,
$\frac{C}{n^t}\leq \frac{C}{n^2}\leq P(x^2\geq n)\leq \frac{Ex^4}{n^2}$. But how to find bound on $Ex^4$?
Suppose to the contrary that
$$ P(x^2 \geq n_i) \geq C/n_i^2 $$ for some infinite sequence $n_1$, $n_2$, $\ldots$, $n_i$, $\ldots$, with $P(x^2 &gt; n_i) &gt; 2P(x^2 &gt; n_{i+1})$. Then we have $$ Var(x) = \int_{x=0}^\infty x^2 d \mu.$$
But now, let's set $n_0 = 0$ and break this integral up into 
$$ Var(x) = \sum_{j=0}^\infty \int_{n_j}^{n_{j+1}}x^2 d \mu.$$
We have$$ \int_{n_j}^{n_{j+1}} x^2 d\mu \geq n_j^2 \int_{n_j}^{n_{j+1}} d\mu = n_j^2(P(x\geq n_j)-P(x \geq n_{j+1})) \geq C/2.$$
This shows that the integral diverges, contradicting the variance being 1.