I have to prove that for all $x,y,z>0$,
$$\left(\frac{x+y}{x+y+z}\right)^{0.5} + \left(\frac{x+z}{x+y+z}\right)^{0.5} + \left(\frac{z+y}{x+y+z}\right)^{0.5} \leq 6^{0.5}$$
using Cauchy-Schwarz inequality? How do I do that ?
I have to define an inner product but I do not what, and what are the vectors?
Something to notice is that $\left(\frac{x+y}{x+y+z}\right)+ \left(\frac{x+z}{x+y+z}\right) + \left(\frac{z+y}{x+y+z}\right)= 2 $. This suggests writing $\left(\frac{x+y}{x+y+z}\right)^{0.5} + \left(\frac{x+z}{x+y+z}\right)^{0.5} + \left(\frac{z+y}{x+y+z}\right)^{0.5} $ as the dot product of $\left(\left(\frac{x+y}{x+y+z}\right)^{0.5},\left(\frac{x+z}{x+y+z}\right)^{0.5},\left(\frac{z+y}{x+y+z}\right)^{0.5}\right)$ with $(1,1,1)$, since the sum of the squares of the three terms will be on the right-hand side. And sure enough, using the Cauchy-Schwarz inequality will immediately give the bound you seek.
