Max/Min Problem

Here's how I would go about it: As the expression is homogeneous, we can set $6x+5y+4z=1$ WLOG. Then CS inequality gives: $$(\overline{3x+4y}+\overline{y+2z}+\overline{2z+3x})(1+1+1)\ge \left(\sqrt{3x+4y}+\sqrt{y+2z}+\sqrt{2z+3x} \right)^2$$

So we get $\sqrt{3x+4y}+\sqrt{y+2z}+\sqrt{2z+3x} \le \sqrt3$ and equality is possible iff $\dfrac{3x+4y}1=\dfrac{y+2z}1=\dfrac{2z+3x}1 \implies x:y:z = 1:3:6$