Question

For `x,y`, and `z` positive real numbers, what is the maximum possible value for \[ \sqrt{\frac{3x+4y}{6x+5y+4z}} + \sqrt{\frac{y+2z}{6x+5y+4z}} + \sqrt{\frac{2z+3x}{6x+5y+4z}}? \]

Answer 1

`sqrt(3)`

Explanation:

Calling

`{(u=3x+4y),(v=y+2z),(w=3x+2z):}`

we have

# \sqrt{\frac{3x+4y}{6x+5y+4z}} + \sqrt{\frac{y+2z}{6x+5y+4z}} + \sqrt{\frac{2z+3x}{6x+5y+4z}} = sqrt(u/(u+v+w))+sqrt(v/(u+v+w))+sqrt(w/(u+v+w)) #

and also

`sqrt(u/(u+v+w))+sqrt(v/(u+v+w))+sqrt(w/(u+v+w)) le 3/sqrt(3)=sqrt(3)`

because if

`s_n = sum_(k=1)^n sqrt(u_k)`

such that `u_k > 0` for `k=1,2,cdots,n` and
`sum_(k=1)^n u_k = 1`

then

`s_n le n/sqrt(n)=sqrt(n)`