Nelson Faustino guessed that $(a^2+b^2)(b^2+c^2)(c^2+a^2)\le 8$. Let us prove it.
Proof: We will use the following lemma whose proof is given later.
Lemma 1: Let $a, b, c\ge 0$ with $11(a^2+b^2+c^2) + 4(a+b+c) - 45 \le 0$. Then
$(a^2+b^2)(b^2+c^2)(c^2+a^2)\le 8$.
Let us begin. Using AM-GM, we have
$$\frac{1}{(a^2+b^2)^2} + \frac{1}{(b^2+c^2)^2} + \frac{1}{(c^2+a^2)^2}
\ge 3\sqrt[3]{\frac{1}{(a^2+b^2)^2(b^2+c^2)^2(c^2+a^2)^2}}.$$
It suffices to prove that $(a^2+b^2)(b^2+c^2)(c^2+a^2)\le 8$.
To this end, we have
\begin{align}
&a, b, c > 0, \quad a^ab^bc^c=1 \qquad\qquad (1)\\
\Longrightarrow \quad & a\ln a + b\ln b + c\ln c = 0, \quad 0 < a, b, c < \frac{159}{100}\qquad\qquad (2)\\
\Longrightarrow \quad &11(a^2+b^2+c^2) + 4(a+b+c) - 45 \le 0, \quad a, b, c > 0.
\qquad\qquad (3)
\end{align}
Here, in (2), since $-u\ln u \le \frac{1}{\mathrm{e}}$ for $u>0$, we have
$a\ln a, b\ln b, c\ln c \le \frac{2}{\mathrm{e}}$ which, when combined with $\frac{159}{100}\ln \frac{159}{100} > \frac{2}{\mathrm{e}}$,
results in $a, b, c < \frac{159}{100}$ (noting that $x \mapsto x\ln x$ is strictly increasing for $x > 1$);
In (3), we have used the fact that
$$x\ln(x) \ge \frac{11}{26}x^2 + \frac{2}{13}x - \frac{15}{26}, \quad \forall 0 < x < \frac{159}{100}.\qquad (4)$$
Then, according to Lemma 1, the desired result follows. We are done.
Proof of (4): It suffices to prove that
$$\ln x \ge \frac{\frac{11}{26}x^2 + \frac{2}{13}x - \frac{15}{26}}{x}, \quad \forall 0 < x < \frac{159}{100}.$$
Let $h(x) = \ln x - \frac{\frac{11}{26}x^2 + \frac{2}{13}x - \frac{15}{26}}{x}$.
We have $h'(x) = -\frac{11(x-1)(x-\frac{15}{11})}{26x^2}$.
Thus, $h(x)$ is decreasing on $(0, 1)$, increasing on $(1, \frac{15}{11})$ and decreasing on $(\frac{15}{11}, \infty)$.
Note that $h(1) = 0$ and $h(\frac{159}{100}) > 0 $. The desired result follows.
Proof of Lemma 1: One of the methods is the uvw method as follows. However, I hope to see nice proofs of Lemma 1.
Let $a+b+c = 3u$, $ab+bc+ca = 3v^2$ and $abc = w^3$.
The constraint $11(a^2+b^2+c^2) + 4(a+b+c) - 45 \le 0$ becomes $11(9u^2 - 6v^2) + 12u - 45 \le 0$.
The constraint does not involve $w^3$.
We need to prove that $f(w^3) = 8 - (a^2+b^2)(b^2+c^2)(c^2+a^2) = w^6+6u(9u^2-6v^2)w^3-81u^2v^4+54v^6+8 \ge 0$.
Since $9u^2 - 6v^2 \ge 0$ and $w^3\ge 0$, $f(w^3)$ is increasing with $w^3$.
Thus, we only need to prove the case when $a=b$ or $c=0$. The rest is not hard and thus omitted.
By AM-GM, we have $\sqrt{ab} + \sqrt{bc} + \sqrt{ca} \le a + b + c = 3$. Thus,
$\mathrm{RHS} \ge \frac{1}{3}$.
By Chebyshev's sum inequality, we have
\begin{align}
a^3 + b^3 + c^3 - (a+b+c) &= (a-1)(a^2 + a) + (b-1)(b^2+b) + (c-1)(c^2+c)\\
&\ge \frac{1}{3}(a-1 + b-1 + c-1)(a^2+a + b^2+b + c^2 + c)\\
& = 0.
\end{align}
Thus, we have $a^3 + b^3 + c^3 \ge a + b + c = 3$. Thus, we have
\begin{align}
a^6 + b^6 + 3c^3 + 4 &= (a^6 + 1) + (b^6 + 1) + 3c^3 + 2\\
&\ge 2a^3 + 2b^3 + 3c^3 + 2\\
&= 2(a^3 + b^3 + c^3) + c^3 + 2\\
&\ge 8 + c^3.
\end{align}
Thus, $\mathrm{LHS} \le \sum_{\mathrm{cyc}} \frac{1}{8 + c^3}$.
It suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{1}{8 + c^3} \le \frac{1}{3}$$
or (tangent line method)
$$\sum_{\mathrm{cyc}} \left(\frac{1}{8 + c^3} - \frac{1}{9} + \frac{1}{27}(c-1) \right) \le 0$$
or
$$\sum_{\mathrm{cyc}} \frac{(c^2-2c-5)(c-1)^2}{27(8 + c^3)} \le 0$$
which is true since $x^2-2x-5 < 0$ for all $0 \le x \le 3$.
Best Answer
It's wrong.
Try $(a,b,c)=(0.47,0.47,0.06).$
In this case $$1-b+b^{\frac{2}{3}}-\sum_{cyc}\frac{a}{\sqrt[3]{a+b}}=-0.00020...$$