Minimising 1/a + 4/b + 9/c on a + b + c = 1

Cauchy-Schwarz.
$$\left(\frac1a+\frac4b+\frac9c\right)(a+b+c)\ge\left(1+2+3\right)^2=36.$$
Since $a+b+c=1$, $\frac1a+\frac4b+\frac9c\ge36$.

Equality. In Cauchy-Schwarz the two vectors are proportional: $\frac1{\sqrt a}:\sqrt a=\frac2{\sqrt b}:\sqrt b=\frac3{\sqrt c}:\sqrt c$, i.e. $a:b:c=1:2:3$. With $a+b+c=1$: $(a,b,c)=\left(\frac16,\frac13,\frac12\right)$, and indeed $6+12+18=36$.

Remark. This is the Engel form: $\sum\frac{x_i^2}{y_i}\ge\frac{(\sum x_i)^2}{\sum y_i}$ with $x=(1,2,3)$ and $y=(a,b,c)$.

Lesson. To find an extremum with Cauchy-Schwarz, *first* guess the equality case ($a:b:c$), which tells you how to choose the vectors.