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?!

WLOG $a\geq b\geq c.$
Let $f(a,b,c)=a+b+c+1/a+/b+1/c.$
For a given value of $a+b+c,$ let $c$ remain constant while $a,b$ vary , subject to the constraint that $a+b$ is constant, so that $a+b+c$ also remains constant.  Then $db/da=-1$ and $d(1/b)/da=1/a^2 .$ So with constant $c$ we have $$df(a,b,c)/da=-1/a^2+1/b^2=(a-b)(a+b)/a^2b^2\geq 0$$ (because $a\geq b$). So we cannot have a minimum of $f(a,b,c)$ for a given value of $a+b+c$ unless $a=b.$ 
Applying this method again, leaving $a$ constant and letting $b,c$ vary, subject to the constraint that $b+c$ is constant, we see also that we cannot have a minimum of $f$ for a  given value of $a+b+c$ unless $b=c.$ 
Therefore for each $S\in (0,3/2]$ we have $$\min \{f(a,b,c): a+b+c=S\}=f(S/3,S/3,S/3)=S+9/S.$$ The least value of $S+9/S$ for $S\in (0,3/2]$, is $15/2,$ which occurs uniquely  at $S= 3/2.$ And as we have seen , this only occurs when $a=b=c=S/3=1/2.$
