Polynomialized Inequality

Using Cauchy-Schwarz Inequality, we have:

\(\begin{array} {} (4\sqrt{a}+6\sqrt{b}+12\sqrt{c})^2 & \le (4^2+6^2+12^2)(a+b+c) \\ & \le 196(4abc) = 784abc \end{array} \)

$\Rightarrow \dfrac{(4\sqrt{a}+6\sqrt{b}+12\sqrt{c})^2}{abc} \le 784\\ \Rightarrow \dfrac{4\sqrt{a}+6\sqrt{b}+12\sqrt{c}}{\sqrt{abc}} \le \sqrt{784} = \boxed{28}$

Let us take x^2=1/bc, y^2=1/ac, z^2=1/ab. We should then find the max. value of the sum S = 4x +6y +12z, having in mind that x^2 +y^2 +z^2= 4 i.e. all the points (x,y,z) form the surface of the sphere with radius r =2. Then S can be considered as the scalar product of two vectors - (4,6,12) and (x,y,z). The max. value will evidently be sqrt (4^2 + 6^2 + 12^2) 2= 14 2=28