I'm trying to find the maximum possible value of the following expression, where $x,y,z > 0$:
$$\sqrt{\frac{3x+4y}{6x+5y+4z}} + \sqrt{\frac{y+2z}{6x+5y+4z}} + \sqrt{\frac{2z+3x}{6x+5y+4z}}$$
No other stipulations are given. I've tried a number of things, and I think that I need to apply the Cauchy-Schwartz Inequality in some way when relating the numerators and denominator of each square root. Other than that, I'm at a loss.
Let $$a=\sqrt{\frac{3x+4y}{6x+5y+4z}},\qquad b=\sqrt{\frac{y+2z}{6x+5y+4z}}, \qquad c=\sqrt{\frac{2z+3x}{6x+5y+4z}},$$ then $a, b, c>0$ and $a^2+b^2+c^2=1$ and we need to find the maximum value for $a+b+c$. This is quickly given by the famous AM-QM inequality $$\frac{a+b+c}3\leq \sqrt{\frac{a^2+b^2+c^2}3},$$ with equality holds for $a=b=c$. Applying this inequality gives that $a+b+c\leq \sqrt3$ with equality holds for $a=b=c=\frac1{\sqrt3}$, which is further equivalent to $y=3x, z=6x$ according to the definition of $a,b,c$.
Therefore, the maximum value of $\sqrt{\frac{3x+4y}{6x+5y+4z}} + \sqrt{\frac{y+2z}{6x+5y+4z}} + \sqrt{\frac{2z+3x}{6x+5y+4z}}$ is $\sqrt3$, and it takes this maximum value when $y=3x, z=6x$.