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$\newcommand{\Ga}{\Gamma}$Let us show that the negation of your first inequality is true, for each $d\ge2$. 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 $1/2,(n-1)/2$, and hence \begin{equation*} \frac{E\|X_n\|_1}{\sqrt n}=f_n:=\frac2{\sqrt\pi}\,\frac{\Ga(n/2)}{\Ga((n-1)/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}=\frac1{n^2-1}\,{\sqrt{\frac{n^5 (n+3)}{n^2+3 n+2}}}>1 \end{equation*} for $n>1$, because \begin{equation} \rho_n^2-1=\frac{6 n^3-2+3n(n-1)}{(n-1)^2 (n+1)^3 (n+2)}>0. \end{equation} Also, it is easy to see that $r_n\to1$ (as $n\to\infty$). So, $r_n<1$ for all $n>1$, that is, ($*$) holds, as claimed.

Concerning the second inequality, Pierre PC showed a comment that $E\|X_n\|_1^2=n+n(n-1)2/\pi$. Hence, the ratio \begin{equation} \frac{E\|X_n\|_1^2}{n}=\frac2\pi+\frac{1-2/\pi}n \end{equation} is decreasing in $n$, which means the the second inequality holds.

Asking multiple questions in one post is not encouraged on MathOverflow, I think.

$\newcommand{\Ga}{\Gamma}$Your first inequality is true, for each $n:=d\ge2$. 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 $1/2,(n-1)/2$, and hence \begin{equation*} \frac{E\|X_n\|_1}{\sqrt n}=f_n:=\frac{\sqrt n}{\sqrt\pi}\,\frac{\Ga(n/2)}{\Ga((n+1)/2)}, \end{equation*} and your first inequality means that \begin{equation*} r_n:=f_n/f_{n+1}\overset{\text{(?)}}\ge1. \tag{$*$} \end{equation*}

Note that \begin{equation*} \rho_n:=\frac{r_{n+2}}{r_n}=\frac{(n+2)^{3/2}}{(n+1)^{3/2}}\,\sqrt{\frac{n}{n+3}}<1 \end{equation*} for $n>0$, because \begin{equation} \rho_n^2-1=-\frac{2 n+3}{(n+1)^3 (n+3)}<0. \end{equation} So, $r_{n+2j}$ is decreasing in $j\in\{0,1,\dots\}$, for each $n>0$. 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 desired.

Concerning the second inequality, Pierre PC showed in a comment that $E\|X_n\|_1^2=1+(n-1)2/\pi$. Hence, the ratio \begin{equation} \frac{E\|X_n\|_1^2}{n}=\frac2\pi+\frac{1-2/\pi}n \end{equation} is decreasing in $n$, which means the the second inequality holds as well.

Asking multiple questions in one post is not encouraged on MathOverflow, I think.

$\newcommand{\Ga}{\Gamma}$Let us show that the negation of your first inequality is true, for each $d\ge2$. 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 $1/2,(n-1)/2$, and hence \begin{equation*} \frac{E\|X_n\|_1}{\sqrt n}=f_n:=\frac2{\sqrt\pi}\,\frac{\Ga(n/2)}{\Ga((n-1)/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}=\frac1{n^2-1}\,{\sqrt{\frac{n^5 (n+3)}{n^2+3 n+2}}}>1 \end{equation*} for $n>1$, because \begin{equation} \rho_n^2-1=\frac{6 n^3-2+3n(n-1)}{(n-1)^2 (n+1)^3 (n+2)}>0. \end{equation} Also, it is easy to see that $r_n\to1$ (as $n\to\infty$). So, $r_n<1$ for all $n>1$, that is, ($*$) holds, as claimed.

The second inequality can apparently be treated similarly, using now the fact the joint distribution of the squares of two coordinates of the random vector $X_n$ is the Dirichlet distribution (of order $3$) with parameters $1/2,1/2,(n-2)/2$. (The beta distribution is a Dirichlet distribution of order $2$.)

Asking multiple questions in one post is not encouraged on MathOverflow, I think.

$\newcommand{\Ga}{\Gamma}$Let us show that the negation of your first inequality is true, for each $d\ge2$. 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 $1/2,(n-1)/2$, and hence \begin{equation*} \frac{E\|X_n\|_1}{\sqrt n}=f_n:=\frac2{\sqrt\pi}\,\frac{\Ga(n/2)}{\Ga((n-1)/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}=\frac1{n^2-1}\,{\sqrt{\frac{n^5 (n+3)}{n^2+3 n+2}}}>1 \end{equation*} for $n>1$, because \begin{equation} \rho_n^2-1=\frac{6 n^3-2+3n(n-1)}{(n-1)^2 (n+1)^3 (n+2)}>0. \end{equation} Also, it is easy to see that $r_n\to1$ (as $n\to\infty$). So, $r_n<1$ for all $n>1$, that is, ($*$) holds, as claimed.

Concerning the second inequality, Pierre PC showed a comment that $E\|X_n\|_1^2=n+n(n-1)2/\pi$. Hence, the ratio \begin{equation} \frac{E\|X_n\|_1^2}{n}=\frac2\pi+\frac{1-2/\pi}n \end{equation} is decreasing in $n$, which means the the second inequality holds.

Asking multiple questions in one post is not encouraged on MathOverflow, I think.

$\newcommand{\Ga}{\Gamma}$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 $1/2,(n-1)/2$, and hence \begin{equation*} \frac{E\|X_n\|_1}{\sqrt n}=f_n:=\frac2{\sqrt\pi}\,\frac{\Ga(n/2)}{\Ga((n-1)/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}=\frac1{n^2-1}\,{\sqrt{\frac{n^5 (n+3)}{n^2+3 n+2}}}>1 \end{equation*} for $n>1$, because \begin{equation} \rho_n^2-1=\frac{6 n^3-2+3n(n-1)}{(n-1)^2 (n+1)^3 (n+2)}>0. \end{equation} Also, it is easy to see that $r_n\to1$ (as $n\to\infty$). So, $r_n<1$ for all $n>1$, that is, ($*$) holds, as claimed.

The second inequality can apparently be treated similarly, using now the fact the joint distribution of the squares of two coordinates of the random vector $X_n$ is the Dirichlet distribution (of order $3$) with parameters $1/2,1/2,(n-2)/2$. (The beta distribution is a Dirichlet distribution of order $2$.)

Asking multiple questions in one post is not encouraged on MathOverflow, I think.

$\newcommand{\Ga}{\Gamma}$Let us show that the negation of your first inequality is true, for each $d\ge2$. 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 $1/2,(n-1)/2$, and hence \begin{equation*} \frac{E\|X_n\|_1}{\sqrt n}=f_n:=\frac2{\sqrt\pi}\,\frac{\Ga(n/2)}{\Ga((n-1)/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}=\frac1{n^2-1}\,{\sqrt{\frac{n^5 (n+3)}{n^2+3 n+2}}}>1 \end{equation*} for $n>1$, because \begin{equation} \rho_n^2-1=\frac{6 n^3-2+3n(n-1)}{(n-1)^2 (n+1)^3 (n+2)}>0. \end{equation} Also, it is easy to see that $r_n\to1$ (as $n\to\infty$). So, $r_n<1$ for all $n>1$, that is, ($*$) holds, as claimed.

The second inequality can apparently be treated similarly, using now the fact the joint distribution of the squares of two coordinates of the random vector $X_n$ is the Dirichlet distribution (of order $3$) with parameters $1/2,1/2,(n-2)/2$. (The beta distribution is a Dirichlet distribution of order $2$.)

Asking multiple questions in one post is not encouraged on MathOverflow, I think.