How does this follow from Chebyshev's inequality My textbook claims that the following inequality follows from chebshev's inequality
$\int_{|x|\geq B}(1+|x|^{p})dL_{A}\leq B^{-p-2q}\int(1+x^{2(p+q)})dL_{A},$
where $L_{A}$ is the empirical spectral disribution of some random matrix, that is of the form $L_{A}=\frac{1}{N}\sum_{i=1}^{N}\delta_{\lambda_{i}}$ where $N$ is the size of the matrix (In fact $L_{A}$ could be any probability measure).
This inequality is not clear for me. I have separated the LHS into a sum of two integrals, for the first integral there is no problem,  for the second I use $g(u)=u^{\frac{2(p+q)}{p}}$ to find
$\int_{|x|\geq B}|x|^{p}dL_{A}\leq \frac{1}{B^{\frac{2(p+q)}{p}}}\int x^{2(p+q)}dL_{A},$
which is not what I want.

$$ \int_{|x|\ge B} |x|^p \, dL_A = \int_{|x|\ge B} |x|^{2(q+p)} |x|^{-p-2q} \, dL_A $$
Since $|x|\ge B$ inside the integral, we get $|x|^{-p-2q} \le B^{-p-2q}$ inside the integral.
