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I add details to Friedrich's comment:

The question easily reduces to the case of a semisimple group. Let $G$ be a quasi-split semisimple group over a field $k$. Let $B\subset G$ be a Borel subgroup defined over $k$, and let $T\subset B$ be a maximal torus. Consider the Dynkin diagram $D=D( G^s,T^s,B^s)$ of our triple over a separable closure $k^s$ of $k$. The absolute Galois group $\Gamma={\rm Gal}(k^s/k)$ naturally acts on $D$. Then the $k$-rank of $G$ is the number of orbits of $\Gamma$ in $D$.

EDIT-1: we provide details. Namely, let $S=S(G^s,T^s,B^s)$ denote the set of simple roots. We may assume that $S$ is the set of vertices of $D$. Let $O\subseteq D$ be an orbit of $\Gamma$ in $S$. For each simple root $\alpha\in O\subseteq S$, let $$\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s$$ denote the corresponding coroot. Consider the cocharacter $$\nu_O=\sum_{\alpha\in O}\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s,\quad\ z\mapsto \prod_{\alpha\in O}\alpha^\vee(z)\ \,\text{for}\ z\in (k^s)^\times.$$

Then the image ${\rm im}\, \nu_O$ is a one-dimensional split subtorus of $T$. We may and shall assume that our semisimple group $G$ is simply connected; then the (direct) product $$\prod_{O\subseteq S} {\rm im}\, \nu_O\subseteq T$$ is a maximal split subtorus of $T$ and a maximal split torus in $G$. Its dimension is the number of orbits $O$ of $\Gamma$ in $S$.

EDIT-2 (suggested by @YCor). I consider a question of OP that I state as follows: Is it true that the $k$-rank of a quasi-split semisimple $k$-group is greater or equal to half of its absolute rank?

I add details to David's comment. Let us assume that $G$ is a non-split quasi-split absolutely simple $k$-group. Let $\Gamma_{\rm eff}$ denote the image of $\Gamma$ in ${\rm Aut}\,D$. Since $G$ is non-split, $\Gamma_{\rm eff}$ is non-trivial. Since $G$ is absolutely simple, $D$ is connected. Thus $D$ is a connected Dynkin diagram admitting non-trivial automorphisms. One can easily see that in the case ${\sf A}_n$ we have $\lceil n/2\rceil$ orbits, in the case ${\sf D}_n$ for $n\ge 5$ we have $n-1$ orbits, in the case ${\sf E}_6$ we have 4 orbits, and in the case ${\sf D}_4$ we have 2 or 3 orbits. Thus the answer is "Yes" for absolutely simple groups: indeed, the $k$-rank of $G$ is greater or equal to half of its absolute rank.

Now we do not assume that $G$ is absolutely simple. We can take $G=R_{K/k}{\rm SL}_{2,K}$ where $K/k$ is a finite separable extension of degree $n$, and $R_{K/k}$ denotes Weil's restriction of scalars. Then the absolute rank of $G$ is $n$, whereas the $k$-rank of $G$ is 1. Thus the answer is "No".

$\DeclareMathOperator\Gal{Gal}\DeclareMathOperator\im{im}\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\SL{SL}$I add details to Friedrich's comment:

The question easily reduces to the case of a semisimple group. Let $G$ be a quasi-split semisimple group over a field $k$. Let $B\subset G$ be a Borel subgroup defined over $k$, and let $T\subset B$ be a maximal torus. Consider the Dynkin diagram $D=D( G^s,T^s,B^s)$ of our triple over a separable closure $k^s$ of $k$. The absolute Galois group $\Gamma=\Gal(k^s/k)$ naturally acts on $D$. Then the $k$-rank of $G$ is the number of orbits of $\Gamma$ in $D$.

EDIT-1: we provide details. Namely, let $S=S(G^s,T^s,B^s)$ denote the set of simple roots. We may assume that $S$ is the set of vertices of $D$. Let $O\subseteq D$ be an orbit of $\Gamma$ in $S$. For each simple root $\alpha\in O\subseteq S$, let $$\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s$$ denote the corresponding coroot. Consider the cocharacter $$\nu_O=\sum_{\alpha\in O}\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s,\quad\ z\mapsto \prod_{\alpha\in O}\alpha^\vee(z)\ \,\text{for}\ z\in (k^s)^\times.$$

Then the image $\im \nu_O$ is a one-dimensional split subtorus of $T$. We may and shall assume that our semisimple group $G$ is simply connected; then the (direct) product $$\prod_{O\subseteq S} \im \nu_O\subseteq T$$ is a maximal split subtorus of $T$ and a maximal split torus in $G$. Its dimension is the number of orbits $O$ of $\Gamma$ in $S$.

EDIT-2 (suggested by @YCor). I consider a question of OP that I state as follows: Is it true that the $k$-rank of a quasi-split semisimple $k$-group is greater or equal to half of its absolute rank?

I add details to David's comment. Let us assume that $G$ is a non-split quasi-split absolutely simple $k$-group. Let $\Gamma_{\rm eff}$ denote the image of $\Gamma$ in $\Aut D$. Since $G$ is non-split, $\Gamma_{\rm eff}$ is non-trivial. Since $G$ is absolutely simple, $D$ is connected. Thus $D$ is a connected Dynkin diagram admitting non-trivial automorphisms. One can easily see that in the case ${\sf A}_n$ we have $\lceil n/2\rceil$ orbits, in the case ${\sf D}_n$ for $n\ge 5$ we have $n-1$ orbits, in the case ${\sf E}_6$ we have 4 orbits, and in the case ${\sf D}_4$ we have 2 or 3 orbits. Thus the answer is "Yes" for absolutely simple groups: indeed, the $k$-rank of $G$ is greater or equal to half of its absolute rank.

Now we do not assume that $G$ is absolutely simple. We can take $G=R_{K/k}\SL_{2,K}$ where $K/k$ is a finite separable extension of degree $n$, and $R_{K/k}$ denotes Weil's restriction of scalars. Then the absolute rank of $G$ is $n$, whereas the $k$-rank of $G$ is 1. Thus the answer is "No".

I add details to Friedrich's comment:

The question easily reduces to the case of a semisimple group. Let $G$ be a quasi-split semisimple group over a field $k$. Let $B\subset G$ be a Borel subgroup defined over $k$, and let $T\subset B$ be a maximal torus. Consider the Dynkin diagram $D=D( G^s,T^s,B^s)$ of our triple over a separable closure $k^s$ of $k$. The absolute Galois group $\Gamma={\rm Gal}(k^s/k)$ naturally acts on $D$. Then the $k$-rank of $G$ is the number of orbits of $\Gamma$ in $D$.

EDIT-1: we provide details. Namely, let $S=S(G^s,T^s,B^s)$ denote the set of simple roots. We may assume that $S$ is the set of vertices of $D$. Let $O\subseteq D$ be an orbit of $\Gamma$ in $S$. For each simple root $\alpha\in O\subseteq S$, let $$\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s$$ denote the corresponding coroot. Consider the cocharacter $$\nu_O=\sum_{\alpha\in O}\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s,\quad\ z\mapsto \prod_{\alpha\in O}\alpha^\vee(z)\ \,\text{for}\ z\in (k^s)^\times.$$

Then the image ${\rm im}\, \nu_O$ is a one-dimensional split subtorus of $T$. We may and shall assume that our semisimple group $G$ is simply connected; then the (direct) product $$\prod_{O\subseteq S} {\rm im}\, \nu_O\subseteq T$$ is a maximal split subtorus of $T$ and a maximal split torus in $G$. Its dimension is the number of orbits $O$ of $\Gamma$ in $S$.

EDIT-2 (suggested by @YCor). I consider a question of OP that I state as follows: Is it true that the $k$-rank of a quasi-split semisimple $k$-group is greater or equal to half of its absolute rank?

I add details to David's comment. Let us assume that $G$ is a non-split quasi-split absolutely simple $k$-group. Let $\Gamma_{\rm eff}$ denote the image of $\Gamma$ in ${\rm Aut}\,D$. Since $G$ is non-split, $\Gamma_{\rm eff}$ is non-trivial. Since $G$ is absolutely simple, $D$ is connected. Thus $D$ is a connected Dynkin diagram admitting non-trivial automorphisms. One can easily see that in the case ${\sf A}_n$ we have $\lceil n/2\rceil$ orbits, in the case ${\sf D}_{2n}$ for $2n\ge 6$ we have $2n-1$ orbits, in the case ${\sf E}_6$ we have 4 orbits, and in the case ${\sf D}_4$ we have 2 or 3 orbits. Thus the answer is "Yes" for absolutely simple groups: indeed, the $k$-rank of $G$ is greater or equal to half of its absolute rank.

Now we do not assume that $G$ is absolutely simple. We can take $G=R_{K/k}{\rm SL}_{2,K}$ where $K/k$ is a finite separable extension of degree $n$, and $R_{K/k}$ denotes Weil's restriction of scalars. Then the absolute rank of $G$ is $n$, whereas the $k$-rank of $G$ is 1. Thus the answer is "No".

I add details to Friedrich's comment:

The question easily reduces to the case of a semisimple group. Let $G$ be a quasi-split semisimple group over a field $k$. Let $B\subset G$ be a Borel subgroup defined over $k$, and let $T\subset B$ be a maximal torus. Consider the Dynkin diagram $D=D( G^s,T^s,B^s)$ of our triple over a separable closure $k^s$ of $k$. The absolute Galois group $\Gamma={\rm Gal}(k^s/k)$ naturally acts on $D$. Then the $k$-rank of $G$ is the number of orbits of $\Gamma$ in $D$.

EDIT-1: we provide details. Namely, let $S=S(G^s,T^s,B^s)$ denote the set of simple roots. We may assume that $S$ is the set of vertices of $D$. Let $O\subseteq D$ be an orbit of $\Gamma$ in $S$. For each simple root $\alpha\in O\subseteq S$, let $$\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s$$ denote the corresponding coroot. Consider the cocharacter $$\nu_O=\sum_{\alpha\in O}\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s,\quad\ z\mapsto \prod_{\alpha\in O}\alpha^\vee(z)\ \,\text{for}\ z\in (k^s)^\times.$$

Then the image ${\rm im}\, \nu_O$ is a one-dimensional split subtorus of $T$. We may and shall assume that our semisimple group $G$ is simply connected; then the (direct) product $$\prod_{O\subseteq S} {\rm im}\, \nu_O\subseteq T$$ is a maximal split subtorus of $T$ and a maximal split torus in $G$. Its dimension is the number of orbits $O$ of $\Gamma$ in $S$.

EDIT-2 (suggested by @YCor). I consider a question of OP that I state as follows: Is it true that the $k$-rank of a quasi-split semisimple $k$-group is greater or equal to half of its absolute rank?

I add details to David's comment. Let us assume that $G$ is a non-split quasi-split absolutely simple $k$-group. Let $\Gamma_{\rm eff}$ denote the image of $\Gamma$ in ${\rm Aut}\,D$. Since $G$ is non-split, $\Gamma_{\rm eff}$ is non-trivial. Since $G$ is absolutely simple, $D$ is connected. Thus $D$ is a connected Dynkin diagram admitting non-trivial automorphisms. One can easily see that in the case ${\sf A}_n$ we have $\lceil n/2\rceil$ orbits, in the case ${\sf D}_n$ for $n\ge 5$ we have $n-1$ orbits, in the case ${\sf E}_6$ we have 4 orbits, and in the case ${\sf D}_4$ we have 2 or 3 orbits. Thus the answer is "Yes" for absolutely simple groups: indeed, the $k$-rank of $G$ is greater or equal to half of its absolute rank.

Now we do not assume that $G$ is absolutely simple. We can take $G=R_{K/k}{\rm SL}_{2,K}$ where $K/k$ is a finite separable extension of degree $n$, and $R_{K/k}$ denotes Weil's restriction of scalars. Then the absolute rank of $G$ is $n$, whereas the $k$-rank of $G$ is 1. Thus the answer is "No".

To add details to Friedrich's comment:

The question easily reduces to the case of a semisimple group. Let $G$ be a quasi-split semisimple group over a field $k$. Let $B\subset G$ be a Borel subgroup defined over $k$, and let $T\subset B$ be a maximal torus. Consider the Dynkin diagram $D=D( G^s,T^s,B^s)$ of our triple over a separable closure $k^s$ of $k$. The absolute Galois group $\Gamma={\rm Gal}(k^s/k)$ naturally acts on $D$. Then the $k$-rank of $G$ is the number of orbits of $\Gamma$ in $D$.

EDIT: we provide details. Namely, let $S=S(G^s,T^s,B^s)$ denote the set of simple roots. We may assume that $S$ is the set of vertices of $D$. Let $O\subseteq D$ be an orbit of $\Gamma$ in $S$. For each simple root $\alpha\in O\subseteq S$, let $$\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s$$ denote the corresponding coroot. Consider the cocharacter $$\nu_O=\sum_{\alpha\in O}\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s,\quad\ z\mapsto \prod_{\alpha\in O}\alpha^\vee(z)\ \,\text{for}\ z\in (k^s)^\times.$$

Then the image ${\rm im}\, \nu_O$ is a one-dimensional split subtorus of $T$. We may and shall assume that our semisimple group $G$ is simply connected; then the (direct) product $$\prod_{O\subseteq S} {\rm im}\, \nu_O\subseteq T$$ is a maximal split subtorus of $T$ and a maximal split torus in $G$. Its dimension is the number of orbits $O$ of $\Gamma$ in $S$.

For example, if $G$ is a quasi-split absolutely simple group of type ${\sf E}_6$, then its $k$-rank is 6 if it is split, and 4 if not. Similarly, a quasi-split absolutely simple group of type ${\sf D}_4$ can have $k$-rank 4, 3, or 2.

I add details to Friedrich's comment:

The question easily reduces to the case of a semisimple group. Let $G$ be a quasi-split semisimple group over a field $k$. Let $B\subset G$ be a Borel subgroup defined over $k$, and let $T\subset B$ be a maximal torus. Consider the Dynkin diagram $D=D( G^s,T^s,B^s)$ of our triple over a separable closure $k^s$ of $k$. The absolute Galois group $\Gamma={\rm Gal}(k^s/k)$ naturally acts on $D$. Then the $k$-rank of $G$ is the number of orbits of $\Gamma$ in $D$.

EDIT-1: we provide details. Namely, let $S=S(G^s,T^s,B^s)$ denote the set of simple roots. We may assume that $S$ is the set of vertices of $D$. Let $O\subseteq D$ be an orbit of $\Gamma$ in $S$. For each simple root $\alpha\in O\subseteq S$, let $$\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s$$ denote the corresponding coroot. Consider the cocharacter $$\nu_O=\sum_{\alpha\in O}\alpha^\vee\colon {\Bbb G}_{{\rm m},k^s}\to T^s,\quad\ z\mapsto \prod_{\alpha\in O}\alpha^\vee(z)\ \,\text{for}\ z\in (k^s)^\times.$$

Then the image ${\rm im}\, \nu_O$ is a one-dimensional split subtorus of $T$. We may and shall assume that our semisimple group $G$ is simply connected; then the (direct) product $$\prod_{O\subseteq S} {\rm im}\, \nu_O\subseteq T$$ is a maximal split subtorus of $T$ and a maximal split torus in $G$. Its dimension is the number of orbits $O$ of $\Gamma$ in $S$.

EDIT-2 (suggested by @YCor). I consider a question of OP that I state as follows: Is it true that the $k$-rank of a quasi-split semisimple $k$-group is greater or equal to half of its absolute rank?

I add details to David's comment. Let us assume that $G$ is a non-split quasi-split absolutely simple $k$-group. Let $\Gamma_{\rm eff}$ denote the image of $\Gamma$ in ${\rm Aut}\,D$. Since $G$ is non-split, $\Gamma_{\rm eff}$ is non-trivial. Since $G$ is absolutely simple, $D$ is connected. Thus $D$ is a connected Dynkin diagram admitting non-trivial automorphisms. One can easily see that in the case ${\sf A}_n$ we have $\lceil n/2\rceil$ orbits, in the case ${\sf D}_{2n}$ for $2n\ge 6$ we have $2n-1$ orbits, in the case ${\sf E}_6$ we have 4 orbits, and in the case ${\sf D}_4$ we have 2 or 3 orbits. Thus the answer is "Yes" for absolutely simple groups: indeed, the $k$-rank of $G$ is greater or equal to half of its absolute rank.

Now we do not assume that $G$ is absolutely simple. We can take $G=R_{K/k}{\rm SL}_{2,K}$ where $K/k$ is a finite separable extension of degree $n$, and $R_{K/k}$ denotes Weil's restriction of scalars. Then the absolute rank of $G$ is $n$, whereas the $k$-rank of $G$ is 1. Thus the answer is "No".