Titu's upside down

By Cauchy-Schwarz Inequality, $(x+\frac{y}{4}+\frac{z}{9})(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \ge (1+\frac{1}{2}+\frac{1}{3})^2=(\frac{11}{6})^2$ Thus, $x+\frac{y}{4}+\frac{z}{9} \ge \frac{(\frac{11}{6})^2}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$ $= \frac{11}{6}$ Therefore, $a+b=6+11=\boxed{17}$

Use Lagrange multipliers to find the minimum value of the formula $f(x,y,z) = x + \frac{y}{4}\ + \frac{z}{9}$ on surface defined by $g(x,y,z) = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{11}{6}$ . Minimum (if it exists) is found when $\nabla f = \lambda\nabla g$ for some value of $\lambda$ . Now $\nabla f = (1,\frac{1}{4},\frac{1}{9})$ and $\nabla g = (\frac{-1}{x^2},\frac{1}{y^2},\frac{1}{z^2})$ , so we see equality at the point (1,2,3) with a value of $\lambda=-1$ . We can search the ray $(t,2t,3t)$ or simply observe that the point (1,2,3) satisfies the constraint. $g(x,y,z) = \frac{11}{6}$ . Plug in the point (1,2,3) to observe minimum $f(1,2,3) = 1 + \frac{2}{4} + \frac{3}{9} = \frac{11}{6}$ in reduced form $\frac{a}{b}$ with $a=11$ and $b=6$ . Thus $a+b=17$ .