Let $a,g$ be positive rational numbers such that $a \gt g$. Let $V$ be the variety of triples of positive numbers whose AM is $a$ and whose GM is $g$ ; in other words,
$$ V=\lbrace (x_1,x_2,x_3) \in {{\mathbb R}_{+}}^3 \ | \ x_1+x_2+x_3=3a, x_1x_2x_3=g^3 \rbrace $$
I have two questions :
(1) Must $V'=V \cap {\mathbb Q}^3$ always be nonempty ?
(2) When $V'$ is nonempty, must it always be dense in $V$ ?
My thoughts : Perhaps one can restrict one's attention to triple of the form $(x_1,x_2,x_3)=(uv^2,u^2w,vw^2)$ with $uvw=g$ : this avoids taking cubic roots.
Working modulo $2$, it is clear that the system $x_1+x_2+x_3=6, x_1x_2x_3=1$ has no solution in $\mathbb Z$ or ${\mathbb F}_2$. I do not know if it has rational solutions or not.
This is not yet a complete answer, but a collection of some partial results:
Observation 1: We may restrict our attention to the case $g=1$.
For positive rational numbers $a>g>0$ let $$V(a,g):=\{ (x_1,x_2,x_3) \in {{\mathbb R}_{+}}^3 \ | \ x_1+x_2+x_3=3a,\ x_1x_2x_3=g^3 \},$$ and let $V'(a,g):=V(a,g)\cap\Bbb{Q}^3$. If $x\in V(a,g)$ then $$x':=\left(\tfrac{x_1}{g},\tfrac{x_2}{g},\tfrac{x_3}{g}\right)\in V'\left(\tfrac ag,1\right),$$ so we may restrict to the case $g=1$, and write $V_a:=V(a,1)$ and $V_a':=V'(a,1)$ for $a>1$.$\quad\square$
Now if $x\in V_a'$ then $x_1x_2x_3=1$ and $x_1+x_2+x_3=3a$, so multiplying by $x_1x_2$ yields $$x_1^2x_2+x_1x_2^2+1=3ax_1x_2.$$ This equation in the variables $x_1$ and $x_2$ defines a plane affine cubic, which homogenizes to the plane projective cubic defined by $$X_1^2X_2+X_1X_2^2+X_3^3=3aX_1X_2X_3.\tag{1}$$ The invertible linear change of variables \begin{eqnarray*} X&:=&12a&X_1+12a&X_2,\\ Y&:=&108a^3&X_1-108a^3&X_2,\\ Z&:=&&X_1+&X_2-3aX_3,\\ \end{eqnarray*} shows that the rational points on $(1)$ correspond bijectively to the rational points on $$Y^2Z=X^3+9a(9a^3-4)X^2Z+432a^2XZ^2-1728a^3Z^3,$$ which is a Weierstrass equation for an elliptic curve. This can be further depressed to the affine $$y^2=x^3-243a^5(9a^3-8)x-27a^3(9a^3-4)(81a^6-81a^5-72a^3+72a^2+16).$$
The following original observations assume that $a\in\Bbb{Z}_+$ is an integer, which yields some interesting results, but they are not as relevant as I had hoped (when I still believed the results held for all rational $a\in\Bbb{Q}_{>0}$).
Let $a\in\Bbb{Z}_{>1}$ be an integer, let $x\in V_a'$ and let $p_i,q_i\in\Bbb{Z}_+$ be such that $x_i=\tfrac{p_i}{q_i}$ and $\gcd(p_i,q_i)=1$.
Observation 2: If $i$, $j$ and $k$ are such that $\{i,j,k\}=\{1,2,3\}$ then $$p_ip_j=q_k^2\qquad\text{ and }\qquad p_iq_i=q_jq_k.$$
Proof. Given that $x\in V_a'$, we can clear denominators to find that $$p_1q_2q_3+q_1p_2q_3+q_1q_2p_3=3aq_1q_2q_3\qquad\text{ and }\qquad p_1p_2p_3=q_1q_2q_3.$$ The former shows that $q_1$ divides $q_2q_3$, so let $c_1:=\tfrac{q_2q_3}{q_1}\in\Bbb{Z}_+$ and define $c_2,c_3\in\Bbb{Z}_+$ similarly. The identities $$p_1p_2p_3=q_1q_2q_3=c_1q_1^2=c_2q_2^2=c_3q_3^2.$$ then show that $p_i\mid c_i$ and hence for $d_i:=\tfrac{c_i}{p_i}\in\Bbb{Z}_+$ we have $$p_2p_3=d_1q_1^2,\qquad p_1p_3=d_2q_2^2\qquad p_1p_2=d_3q_3^2,$$ and then taking the product yields $$(p_1p_2p_3)^2=(p_2p_3)(p_1p_3)(p_1p_2)=(d_1q_1^2)(d_2q_2^2)(d_3q_3^2)=(d_1d_2d_3)(q_1q_2q_3)^2,$$ which shows that $d_1d_2d_3=1$ and hence $d_1=d_2=d_3=1$. We conclude that $$p_1p_2=q_3^2,\qquad p_1p_3=q_2^2,\qquad p_2p_3=q_1^2,$$ from which it follows that $p_1q_1^2=p_1p_2p_3=q_1q_2q_3$, and so $$p_1q_1=q_2q_3,\qquad p_2q_2=q_1q_3,\qquad p_3q_3=q_1q_2.\qquad\square$$
Note also that $\gcd(p_i,q_i)=1$ and $p_1p_2p_3=q_1q_2q_3$ implies $\gcd(q_1,q_2,q_3)=1$.
Observation 3: There exist pairwise coprime positive integers $r_1,r_2,r_3\in\Bbb{Z}_+$ such that $$p_i=r_i^2\qquad\text{ and }\qquad q_i=r_jr_k.$$
Proof. Set $r_i:=\gcd(q_j,q_k)$. Then $\gcd(r_i,r_j)=\gcd(q_1,q_2,q_3)=1$, so $r_1$, $r_2$ and $r_3$ are pairwise coprime.
Let $\ell$ be a prime number dividing $q_i$. Then $\ell\nmid p_i$ and hence $$v_{\ell}(q_i)=v_{\ell}(p_i)+v_{\ell}(q_i)=v_{\ell}(p_iq_i)=v_{\ell}(q_jq_k)=v_{\ell}(q_j)+v_{\ell}(q_k).$$ As either $v_{\ell}(q_j)=0$ or $v_{\ell}(q_k)=0$ we see that either $v_{\ell}(q_j)=v_{\ell}(q_i)$ or $v_{\ell}(q_k)=v_{\ell}(q_i)$, respectively, so either $$v_{\ell}(r_k)=v_{\ell}(q_i)\qquad\text{ or }\qquad v_{\ell}(r_j)=v_{\ell}(q_i),$$ respectively. Because $r_i$ and $r_j$ are coprime it follows that $q_i=r_jr_k$. Then $$p_1p_2p_3=q_1q_2q_3=(r_1r_2r_3)^2,$$ and because $p_i$ is coprime to $q_i$ it follows that $p_i=r_i^2$.$\quad\square$
It follows that $$3aq_1q_2q_3=p_1q_2q_3+q_1p_2q_3+q_1q_2p_3=(r_1r_2r_3)(r_1^3+r_2^3+r_3^3).$$ and so $$a=\frac{r_1^3+r_2^3+r_3^3}{3r_1r_2r_3},$$ so now one might wonder: