Question

For x, y, and z positive real numbers, what is the maximum possible value for

\(\sqrt{\frac{3x+4y}{6x+5y+4z}}
+
\sqrt{\frac{y+2z}{6x+5y+4z}}
+
\sqrt{\frac{2z+3x}{6x+5y+4z}}?
\)

Also, find what z/x is if (x,y,z) achieves the maximum value.

Thank you! I am so stuck right now.

Answer 1

Let \(a=\sqrt{\frac{3x+4y}{6x+5y+4z}}\\
b=\sqrt{\frac{y+2z}{6x+5y+4z}}\\
c=\sqrt{\frac{2z+3x}{6x+5y+4z}}\)

Firstly, notice that \(a^2+b^2+c^2=1\)

Answer 2

appealing to cauchy schwarz inequality gives
\[a*1+b*1+c*1\le \sqrt{(a^2+b^2+c^2)(1^2+1^2+1^2)}=\sqrt{3}\]