<s>[INST] 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?!
 [/INST] your inequality is equivalent to
$$2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+2(a+b+c)\geq 15$$
By $AM-HM$ we get
$$a+b+c\geq \frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ thus $$2(a+b+c)\geq \frac{18}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ thus we have
$$2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\frac{18}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\geq 15$$
Setting $$t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$ so we have to prove
$$2t+\frac{18}{t}\geq 15$$ this is equivalent to
$$2t^2-15t+18\geq 0$$ or $$t\le \frac{3}{2}$$ or $$t\geq 6$$
or we can consider the function $$f(t)=2t+\frac{18}{t}$$ and $$f'(t)=2-\frac{18}{t^2}$$ and $$f'(t)=0$$ for $t=3$
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