+400 Let x, y, and z be positive real numbers. Find the largest possible value of
sqrt((3x + 4y)/(6x + 5y + 4z)) + sqrt((y + 2z)/(5x + 6y + 4z)) + sqrt((2z + 3x)/(x + 2y + 3z))
+297 \(\sqrt{\frac{3x + 4y}{6x + 5y + 4z}} + \sqrt{\frac{y + 2z}{5x + 6y + 4z}} + \sqrt{\frac{2z + 3x}{x + 2y + 3z}}\)
So, largest value of that.
\(u=3x+4y,v=y+2z,w=3x+2z\)
we have
\(\sqrt{\frac{u}{u+v+w}}+\sqrt{\frac{v}{u+v+w}}+\sqrt{\frac{w}{u+v+w}}\)
and also
\(\sqrt{\frac{u}{u+v+w}}+\sqrt{\frac{v}{u+v+w}}+\sqrt{\frac{w}{u+v+w}}\le\sqrt{3}\)
You can google why. I can't write it in latex since idk how.
Answer: \({\sqrt{3}}\)
