Note that since $\dfrac{1}{x^2}$ is a decreasing function, $\int_{j-1}^j\frac{\mathrm{d}x}{x^2}>\frac{1}{j^2}$. Therefore,
$$
\int_{j-1}^\infty\frac{\mathrm{d}x}{x^2}>\sum_{n=j}^\infty\frac{1}{n^2}\tag{1}
$$
However to get the inequality cited in the question, we need to evaluate a bit more carefully:
$$
\begin{align}
\int_{j-1/2}^{j+1/2}\frac{\mathrm{d}x}{x^2}
&=\frac{1}{j-1/2}-\frac{1}{j+1/2}\\
&=\frac{1}{j^2-1/4}\\
&>\frac{1}{j^2}\tag{2}
\end{align}
$$
Summing $(2)$ yields
$$
\int_{j-1/2}^\infty\frac{\mathrm{d}x}{x^2}>\sum_{n=j}^\infty\frac{1}{n^2}\tag{3}
$$
Using the strict convexity of $\mathbf{\dfrac{1}{x^2}}$
We can prove $(3)$ using strict convexity. Since $x$ is not constant on the support of uniform measure ($\mathrm{d}x$) on $[j{-}\frac12,j{+}\frac12]$, Jensen's inequality yields
$$
\int_{j-1/2}^{j+1/2}\frac{\mathrm{d}x}{x^2}>\frac{1}{\left(\int_{j-1/2}^{j+1/2}x\,\mathrm{d}x\right)^2}=\frac{1}{j^2}
$$