Solve this with CBS: minimum value of $ 1/x + 4/y + 9/z $ with $x+y+z=1$ How can you see the minimum value of $ 1/x + 4/y + 9/z $ with $x+y+z=1$ using the CBS inequality?   
I have seen a proof of that that use trigonometric substitutions, but I don't see as one-step the solution using the CBS inequality.

Consider positive values $x,y,z$.
Denote 
$$(a,b,c)=\left( \frac{1}{\sqrt{x}}, \frac{2}{\sqrt{y}}, \frac{3}{\sqrt{z}}\right),$$
$$(A,B,C)=\left( \sqrt{x}, \sqrt{y}, \sqrt{z}\right),$$
then (by CBS inequality)
$$(a^2+b^2+c^2)(A^2+B^2+C^2)\ge (aA+bB+cC)^2,$$
so
$$
\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right) \left(x+y+z\right)\ge (1+2+3)^2=36;
$$
$$
\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\ge 36.
 $$ (I have added the square)
