Being order of variance leads to order of standard deviation?

Is it true to say: $$a = o\Big(\mathrm{Var}(X)\Big) \implies a = o\Big(\sqrt{\operatorname{Var}(X)}\Big)$$ AS I think by the definition of $$o$$ notation we can have: \begin{align} a = o\Big(\mathrm{Var}(X)\Big) &\implies \frac{a}{\mathrm{Var}(X)} \rightarrow 0 \quad\text{and}\quad \operatorname{Var}(X)>\sqrt{\operatorname{Var}(X)}&\\ &\implies \frac{a}{\sqrt{\operatorname{Var}(X)}} \rightarrow0 &\\ &\implies a = o\Big(\sqrt{\operatorname{Var}(X)}\Big)& \end{align} Am I thinking right? any why?

I am not sure this exactly corresponds to your question, but here is a result that may be useful in that context

If $E(z_n^2)=O(a_n^2)$, $$z_n=O_p(a_n).$$

Proof: By assumption there is a $M_1$ and $n_\epsilon$, so that $$E(z_n^2)<M_1^2a_n^2$$ for $n\geqslant n_\epsilon$. From the Tschebychev inequality it follows that, for arbitrary $M_2>0$, $$\Pr(|z_n|\geqslant M_2a_n)\leqslant\frac{E(z_n^2)}{M_2^2a_n^2}$$ Now choose, for some $\epsilon>0$, $M_2\geqslant M_1\epsilon^{-1/2}$, so that \begin{eqnarray*} \Pr(|z_n|/a_n\geqslant M_2)&<&\frac{M_1^2a_n^2}{M_2^2a_n^2}\\ &\leqslant&\frac{M_1^2}{M_1^2\epsilon^{-1}}\\ &=&\epsilon \end{eqnarray*}

Remark:

If we have $$E(z_n-E(z_n))^2=O(a_n^2)$$ and $$E(z_n)=O(a_n),$$ then $$E(z_n^2)=E(z_n-E(z_n))^2+E(z_n)^2$$ implies that $$z_n=O_p(a_n).$$ The rate of the sequence of r.v.s then corresponds to the rate of the standard deviation.