Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ 
Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$ ($a,b,c$ are positive real numbers).

There is a solution, which relies on guessing the minimum case happening at $a=b=c=\frac12$ and then applying AM-GM inequality,but what if one CANNOT guess that?!

Hint:  by the AM-HM (arithmetic-harmonic mean) inequality:
$$
\frac{a+b+c}{3} \ge \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \;\;\iff\;\; \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \ge \frac{9}{a+b+c}
$$
Let $x=a+b+c \in (0,\frac{3}{2}]\,$, then the expression to be minimized can be written as:
$$
a+b+c+\frac1a+\frac1b+\frac1c \ge a+b+c+\frac{9}{a+b+c} = x + \frac{9}{x}
$$
The function $f(x)=x + \frac{9}{x}$ is decreasing on $(0,\frac{3}{2}]\,$, so $f(x) \ge f(\frac{3}{2})=\frac{15}{2}\,$ for $x \in (0,\frac{3}{2}]$.
The minimum value of $\frac{15}{2}$ is attained when $x=\frac{3}{2}$ and AM=HM i.e. $a=b=c=\frac{x}{3}=\frac{1}{2}\,$.
