$\newcommand{\Ga}{\Gamma}$Let us show that the negation of your first inequality is true. Let $n:=d$. Note that $a$ and $b$ equal, respectively, $X_{n+1}$ and $X_n$ in distribution, where \begin{equation*} X_n:=G/|G|, \end{equation*} $G=(G_1,\dots,G_n)$ is a standard Gaussian random vector in $\mathbb R^n$, and $|G|$ is the Euclidean norm of $G$. So, \begin{equation*} E\|X_n\|_1=n\,EY, \end{equation*} where $Y:=|G_1|/|G|$, so that $Y^2$ has the beta distribution with parameters $1/2,n/2$, and hence \begin{equation*} \frac{E\|X_n\|_1}{\sqrt n}=f_n:=\frac2{\sqrt\pi}\,\frac{\Ga((n+1)/2)}{\Ga(n/2)\sqrt n}, \end{equation*} and the negation of your first inequality means that \begin{equation*} r_n:=f_n/f_{n+1}\overset{\text{(?)}}<1. \tag{$*$} \end{equation*}
Note that \begin{equation*} \rho_n:=\frac{r_{n+2}}{r_n}=\frac{(n+1)^{3/2}\,\sqrt{n+3} }{(n+2)^{3/2}\,\sqrt n}>1 \end{equation*} for $n>0$. Also, it is easy to see that $r_n\to1$ (as $n\to\infty$). So, $r_n<1$ for all $n>0$, that is, ($*$) holds, as claimed.