The answer is no. 

Let $x:=\xi_1$ and $y:=\xi_2$, 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 
$$p_t(x,y):=1-P(X_1\le t,X_2\le t).$$
We want to show that 
$$p_t(x,y)\le p_t(1/3,1/3) \tag{1}$$
for all $x,y$ as above and all real $t>0$. 

However, inequality (1) fails to hold for small enough $t>0$ (say for $t=1/10$) if $x=y<1/3$ and $x$ is close enough to $1/3$. 

Details are in a Mathematica notebook whose image is given below: 

[![enter image description here][1]][1]


  [1]: https://i.sstatic.net/2wIzI.png