Let x, y, z be positive real numbers such that x + y + z = 1. The minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$ is R, then R is divisible by 

Clearly, z is a real number in the interval [0, 1]. Hence there is an angle a such that $z={\sin }^{2}a.$ Then x + y = 1 , $x+y=1-{\sin }^{2}a={\cos }^{2}a,$ or $\frac{x}{{\cos }^{2}a}+\frac{y}{{\cos }^{2}a}=1.$ For an angle b, we have ${\cos }^{2}b+{\sin }^{2}b=1.$ Hence, we can set $x={\cos }^{2}a{\cos }^{2}b,y={\cos }^{2}a{\sin }^{2}b$ for some angle b. It suffices to find the minimum value of $$\begin{gathered} P={\sec }^{2}a{\sec }^{2}b+4{\sec }^{2}a{\csc }^{2}b+9{\csc }^{2}a \\ P=({\tan }^2\alpha +1)({\tan }^{2}b+1)+4({\tan }^{2}a+1)({\cot }^{2}b+1)+9({\cot }^{2}a+1) \end{gathered}$$

Expanding the right-hand side gives

$$\begin{gathered} P=14+5{\tan }^{2}a+9{\cot }^{2}a+({\tan }^{2}b+4{\cot }^{2}b)(1+{\tan }^{2}a) \\ \ge 14+5{\tan }^{2}a+9{\cot }^{2}a+2\tan b.2\cot b(1+{\tan }^{2}a) \\ =18+9({\tan }^{2}a+{\cot }^{2}a)\ge 18+9.2\tan a\cot a=36 \end{gathered}$$

Equality holds when $\tan a=\cot a$ and $b=2\cot b,$ which implies that ${\cos }^{2}a={\sin }^{2}a$ and $2{\cos }^{2}b={\sin }^{2}b.$ Because ${\sin }^2\theta +{\cos }^2\theta =1,$ equality holds when ${\cos }^{2}a=\frac{1}{2}$ and ${\cos }^{2}b=\frac{1}{3};$ that is, $x=\frac{1}{6},y=\frac{1}{3},z=\frac{1}{2}.$