+28 Find z/x if (x,y,z) is a triple that gives the maximum value in Part a.
Here is the problem and solution from part A:
Problem:
For positive real numbers x, y, and z, what is the maximum possible value for \(\sqrt{\frac{3x+4y}{6x+5y+4z}} + \sqrt{\frac{y+2z}{6x+5y+4z}} + \sqrt{\frac{2z+3x}{6x+5y+4z}}?\)
Solution:
Notice that the sum of the numerators in the radicals is equal to their common denominator. Therefore we will use Cauchy-Schwarz with the triples
\( \left( \sqrt{\frac{3x+4y}{6x+5y+4z}}, \sqrt{\frac{y+2z}{6x+5y+4z}}, \sqrt{\frac{2z+3x}{6x+5y+4z}} \right) \)
and \((1,1,1)\).
Cauchy-Schwarz tells us
\(\left(\sqrt{\frac{3x+4y}{6x+5y+4z}} + \sqrt{\frac{y+2z}{6x+5y+4z}} + \sqrt{\frac{2z+3x}{6x+5y+4z}}\right)^2 \\ \le (1^2 + 1^2 + 1^2) \left( \frac{3x+4y}{6x+5y+4z} + \frac{y+2z}{6x+5y+4z} + \frac{2z+3x}{6x+5y+4z} \right) \\ = 3 \cdot \frac{6x + 5y + 4z}{6x + 5y + 4z} \\ = 3. \)
Therefore,
\(\sqrt{\frac{3x+4y}{6x+5y+4z}} + \sqrt{\frac{y+2z}{6x+5y+4z}} + \sqrt{\frac{2z+3x}{6x+5y+4z}} \le \boxed{\sqrt{3}}.\)
