Let $x:=\xi_1$ and $y:=\xi_1$, so that $\xi_3=1-x-y$, $0\le x\le y\le1-x-y$, $$m:=\mu_1=\sqrt{\frac{x y}{x+y}},$$ $$r:=\rho=\sqrt{\frac{x(1-x-y)}{(x+y)(1-x)}}.$$ Let also $$p_t(x,y):=P_t(m,r):=1-P(X_1\le t,X_2\le t).$$
Note that $P_t(m,r)$ is increasing in $m$, because an increase of $m$ by $h$ can be modeled by replacing $X_1$ by $X_1+h$.
Also, by Slepian's lemma, $P_t(m,r)$ is increasing in $r$.
Fix any real $t$. Let now $(x,y)$ be a point of maximum of $p_t(x,y)$ over all $(x,y)$ subject to the conditions $0\le x\le y\le1-x-y$. These conditions imply $0\le x\le1/3$. Moreover, if $x=1/3$, the conditions $0\le x\le y\le1-x-y$ imply $y=1/3$, and we are done.
Finally, if $x\ne1/3$ (that is, if $x<1/3$), then there always exists a vector $v$ of the form $(1,B)$ for a real $B$ such that the directional derivatives of both $m$ and $r$ in the direction of $v$ are $>0$, and small enough movements in this direction preserve the conditions $0\le x\le y\le1-x-y$. See details of these calculations with Mathematica in the image below (click on the image to enlarge it). So, such a movement will increase the value of $p_t$, which contradicts the assumption that $(x,y)$ is a point of maximum.
Thus, the only point of maximum is $(x,y)=(1/3,1/3)$, as desired.
