Minimize $\left(a+\frac1a\right)\left(a+\frac1b\right)$+$\left(b+\frac1b\right)\left(b+\frac1c\right)$+$\left(c+\frac1c\right)\left(c+\frac1a\right)$

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.