Minimize $\left(a+\frac1a\right)\left(a+\frac1b\right)$+$\left(b+\frac1b\right)\left(b+\frac1c\right)$+$\left(c+\frac1c\right)\left(c+\frac1a\right)$ Question: Minimize $\left(a+\frac1a\right)\left(a+\frac1b\right)+\left(b+\frac1b\right)\left(b+\frac1c\right)+\left(c+\frac1c\right)\left(c+\frac1a\right)$
when $a+b=c+2$ and $a,b,c>0$.
I tried to replace $c$ with $a+b-2$ but it didn't bring me anywhere..
 A: Let:
$$f(a,b,c):=\left(a+\frac1a\right)\left(a+\frac1b\right)+\left(b+\frac1b\right)\left(b+\frac1c\right)+\left(c+\frac1c\right)\left(c+\frac1a\right)$$
$$f(a,b,c)=\begin{cases}
a,b,c>0\\
a+b=c+2
\end{cases}$$
Wolfram's answer is ${a\rightarrow1.42163, b \rightarrow 1.36931, c \rightarrow 0.790944}$ with $f(a,b,c)=13.1738$. 
This is actually moot, but my answer, if we disregard the second restriction (i.e. $a+b=c+2$), would be $f(a,b,c)=12$ with $a,b,c=1$.
We take the partial derivatives of $f$ w.r.t $a,b,c$ and solve the system of equations. Thus:
$$\begin{align}
\frac{\partial f}{\partial a}=0, \ \frac{\partial f}{\partial b}=0, \ \frac{\partial f}{\partial c}=0,
\end{align}$$
And thus we solve:
$$\begin{align}
\frac{\partial f}{\partial a}&=-\frac{1}{a^2 b}-\frac{c}{a^2}-\frac{1}{a^2 c}+2 a+\frac{1}{b}=0\\
\frac{\partial f}{\partial b}&=-\frac{a}{b^2}-\frac{1}{a b^2}-\frac{1}{b^2 c}+2 b+\frac{1}{c}=0\\
\frac{\partial f}{\partial c}&=-\frac{1}{a c^2}+\frac{1}{a}-\frac{b}{c^2}-\frac{1}{b c^2}+2 c=0,
\end{align}$$
Which gives us a solution over $\mathbb{R}$ : $a,b,c=\begin{cases}1\\-1\end{cases}$, and given the first restriction, $a,b,c=1$ 
It puzzles me that if we take the second restriction into account, $\begin{align}
\frac{\partial f}{\partial a}=0, \ \frac{\partial f}{\partial b}=0, \ \frac{\partial f}{\partial c}=0,
\end{align}$ has $\emptyset$ as a solution.
