<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] For $a=b=c=\frac{1}{2}$ we get a value $\frac{15}{2}$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\left(a+\frac{1}{a}-\frac{5}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-2)(2a-1)}{a}\geq0$$ or
$$\sum_{cyc}\left(\frac{(2a-1)(a-2)}{a}+3(2a-1)\right)+6\left(\frac{3}{2}-a-b-c\right)\geq0$$ or
$$\sum_{cyc}\frac{2(2a-1)^2}{a}+6\left(\frac{3}{2}-a-b-c\right)\geq0,$$
which is obvious.
Done!
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