= Solution
Suppose first that $x\ne0$, and let $E$ be the real span of the <root system>, of dimension equal to the <rank of a semisimple Lie algebra> $\ell$. The images of the roots span $E/\mathbb R\alpha$, so select roots $\beta_1,\ldots,\beta_{\ell-1}$ whose images form a basis of this quotient.
For each $i$, take the highest endpoint $\gamma_i^+$ of the $\alpha$ <root string> through $\beta_i$, and the highest endpoint $\gamma_i^-$ of the <root string> through $-\beta_i$. By construction, $\gamma_i^\pm+\alpha$ is not a root. Moreover $\gamma_i^\pm\ne-\alpha$, since their images in $E/\mathbb R\alpha$ are nonzero. The <root-space decomposition> and its bracket rule therefore give
$$
[\mathfrak g_{\gamma_i^\pm},x]=0.
$$
These $2(\ell-1)$ roots are distinct: their quotient images are the two signs of a basis, which are distinct. Their one-dimensional <root spaces> consequently contribute $2(\ell-1)$ independent vectors to the <Lie algebra centralizer>.
In addition, the $\ell-1$ dimensional subspace $\ker\alpha\subset\mathfrak t$ commutes with $x$, because $[h,x]=\alpha(h)x$. The line $\mathbb Cx\subset\mathfrak g_\alpha$ also commutes with $x$. These contributions are independent by the <root-space decomposition>; in particular none of the selected endpoint roots is $\alpha$. Thus the <centralizer lower bound for a root vector> is
$$
\boxed{\dim Z_{\mathfrak g}(x)\geq2(\ell-1)+(\ell-1)+1=3\ell-2.}
$$
If $x=0$, then $Z_{\mathfrak g}(x)=\mathfrak g$. The <root system> contains the $2\ell$ distinct roots $\pm\alpha_i$, so $\dim\mathfrak g=\ell+|\Phi|\geq3\ell$, which also proves the desired inequality. The argument includes rank one, where the list of $\beta_i$ is empty.
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