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

1 Answer
May 21, 2017

#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)#