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.
The hint.
Rewrite our inequality in the following form.
$$\sum_{cyc}be^a\leq\sqrt[3]e.$$
Show that for any $0<x<1$ the following inequality holds:
$$e^x\leq\frac{9}{4}\left(e+\sqrt[3]e-4\right)x^3+\frac{3}{2}\left(10-e-5\sqrt[3]e\right)x^2+\frac{1}{4}\left(e+21\sqrt[3]e-28\right)x+1.$$
After this it's enough to prove that:
$$\frac{9}{4}\left(e+\sqrt[3]e-4\right)\sum_{cyc}a^3b+\frac{3}{2}\left(10-e-5\sqrt[3]e\right)\sum_{cyc}a^2b+\frac{1}{4}\left(e+21\sqrt[3]e-28\right)\sum_{cyc}ab+1\leq\sqrt[3]e,$$ which after homogenization gives:
$$\sqrt[3]e\sum_{cyc}(4a^4+16a^3b-5a^3c+12a^2b^2-27a^2bc)\geq$$
$$\geq 4\sum_{cyc}(a^4+3a^3b-3a^3c+7a^2b^2-8a^2bc)+e\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc).$$
Now, easy to see that $$\sum_{cyc}(4a^4+16a^3b-5a^3c+12a^2b^2-27a^2bc)\geq0,$$
$$\sum_{cyc}(a^4+3a^3b-3a^3c+7a^2b^2-8a^2bc)\geq0,$$
$$\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)\geq0,$$
$$\sqrt[3]e>1.36$$ and $$e<2.72.$$
Thus, it's enough to prove that:
$$1.36\sum_{cyc}(4a^4+16a^3b-5a^3c+12a^2b^2-27a^2bc)\geq$$
$$\geq 4\sum_{cyc}(a^4+3a^3b-3a^3c+7a^2b^2-8a^2bc)+2.72\sum_{cyc}(4a^3b+a^3c-4a^2b^2-a^2bc)$$ or
$$\sum_{cyc}(18a^4-14a^3b+31a^3c-10a^2b^2-25a^2bc)\geq0$$ and since by Rearrangement
$$\sum_{cyc}a^3c=abc\sum_{cyc}\left(a^2\cdot\frac{1}{b}\right)\geq abc\sum_{cyc}\left(a^2\cdot\frac{1}{a}\right)=\sum_{cyc}a^2bc,$$ it's enough to prove that
$$\sum_{cyc}(18a^4-14a^3b+6a^3c-10a^2b^2)\geq0$$ or
$$\sum_{cyc}(9a^4-7a^3b-5a^2b^2+3ab^3)\geq0$$ or
$$\sum_{cyc}a(a-b)(9a^2+2ab-3b^2)\geq0$$ or
$$\sum_{cyc}\left(a(a-b)(9a^2+2ab-3b^2)-2(a^4-b^4)\right)\geq0$$ or
$$\sum_{cyc}(a-b)^2(7a^2+7ab+2b^2)\geq0$$ and we are done!
Best Answer
Note that (tangent line trick) $$\frac{5+2y}{49} - \frac{y^2}{4y^3+3} = \frac{(8y^2+36y+15)(y-1)^2}{49(4y^3+3)}.$$ Thus, we have $$\frac{y^2}{4y^3+3} \le \frac{5+2y}{49}, \quad \forall y \ge 0.$$ Thus, we have $$\sum_{\mathrm{cyc}} \frac{xy^2}{4y^3+3} \le \sum_{\mathrm{cyc}} \frac{x(5+2y)}{49} = \frac{5(x+y+z) + 2(xy+yz+zx)}{49} \le \frac{3}{7}$$ where we have used the fact that $xy+yz+zx \le \frac{(x+y+z)^2}{3}$. We are done.