# Finding the maximum value of$\sqrt{\frac{3x+4y}{6x+5y+4z}} + \sqrt{\frac{y+2z}{6x+5y+4z}} + \sqrt{\frac{2z+3x}{6x+5y+4z}}?$

1. For $x$, $y$, and $z$ positive real numbers, what is the maximum possible value for

$$\sqrt{\frac{3x+4y}{6x+5y+4z}} + \sqrt{\frac{y+2z}{6x+5y+4z}} + \sqrt{\frac{2z+3x}{6x+5y+4z}}?$$ 2. Find $z/x$ if $(x,y,z)$ achieves the maximum value from Problem 1.

I know I should include the Cauchy Schwarz Inequality in here but I don't know how. And how would I benefit me? Could someone please walk me step by step thorugh these two problems? Thanks!

Just make a change of variables setting $u=3x+4y, v=y+2z, w=2z+3x$, then apply the Cauchy Schwarz inequality to prove $$\sqrt{\frac{u}{u+v+w}}+\sqrt{\frac{v}{u+v+w}}+\sqrt{\frac{w}{u+v+w}}\leq \color{red}{\sqrt{3}}$$ with equality achieved at $(u,v,w)=\lambda(1,1,1)$, than means $(x,y,z)=\lambda(1,3,\color{red}{6})$.