After rearranging terms this is equivalent to showing that $ \left(\frac {b(a + c)}{a + b}\right)^{ab}\left(\frac{c(a + b)}{b + c}\right)^{bc}\left(\frac{a(b + c)}{a + c}\right)^{ca} \leq 1 $ By weighted AM - GM it suffices to show that $ \sum_{cyc}(b(a + c))\left(\frac{ab}{a + b}\right) \leq ab + bc + ca $ Expanding denominators and homogenizing it suffices to show that $ (a + b + c)\sum_{cyc}ab^2(a + c)^2(b + c) \leq (ab + bc + c)^2(a + b)(b + c)(c + a) $. By expanding and cancelling similar terms this is equivalent to $ a^3b^2c^2 + b^3c^2a^2 + c^3a^2b^2 \leq a^4b^2c + b^4c^2a + c^4a^2b $ Dividing both sides by $ a^2b^2c^2 $ this becomes $\frac{a^2}{c} + \frac{b^2}{a} + \frac{c^2}{b} \geq a + b + c $ which is trivial by Cauchy-Schwartz.

Let me add another nice solution from the official solution: Note $f(x)=x\log x$ is convex. The key step: weighted Popoviciu gives \[bf(a)+cf(b)+af(c)+(a+b+c)f\left(\frac{bc+ca+ab}{a+b+c}\right)\ge \sum_{cyc} (b+c)f\left(\frac{ab+bc}{b+c}\right).\]Exponentiating gives \begin{align*} a^{bc}\cdot b^{ca}\cdot c^{ab} \cdot \left(\frac{bc+ca+ab}{a+b+c}\right)^{bc+ca+ab}&\ge \prod_{cyc} \left(\frac{b(a+c)}{b+c}\right)^{bc+ab} \\ &=\prod_{cyc} a^{ab+ca}(b+c)^{ab+ca-bc-ab} \end{align*}Cancelling some terms and using $\frac{bc+ca+ab}{a+b+c}=1$ gives \[1\ge \prod_{cyc} a^{ca}(a+b)^{bc-ab}\]which rearranges to the result. $\quad \blacksquare$