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

Question

3675 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-21T23:34:54

$sqrt(3)$

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

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

For x,yx,y, and zz 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}}? \sqrt{\frac{3x+4y}{6x+5y+4z}} + \sqrt{\frac{y+2z}{6x+5y+4z}} + \sqrt{\frac{2z+3x}{6x+5y+4z}}?