Maximum

$\sqrt{\frac{3x + 4y}{6x + 5y + 4z}} + \sqrt{\frac{y + 2z}{5x + 6y + 4z}} + \sqrt{\frac{2z + 3x}{x + 2y + 3z}}$

So, largest value of that.

$u=3x+4y,v=y+2z,w=3x+2z$

we have

$\sqrt{\frac{u}{u+v+w}}+\sqrt{\frac{v}{u+v+w}}+\sqrt{\frac{w}{u+v+w}}$

and also

$\sqrt{\frac{u}{u+v+w}}+\sqrt{\frac{v}{u+v+w}}+\sqrt{\frac{w}{u+v+w}}\le\sqrt{3}$

You can google why. I can't write it in latex since idk how.

Answer: ${\sqrt{3}}$