Prove that $a^2b+b^2c+c^2a \leqslant 3$ for $a,b,c >0$ with $a^ab^bc^c=1$ 
Let $a,b,c >0$ and $a^ab^bc^c=1$. Prove that
$$a^2b+b^2c+c^2a \leqslant 3.$$

I don't even know what to do with the condition $a^ab^bc^c=1$. At first I think $x^x>1$, but I was wrong. This inequality is true, following by the verification from Mathematica

Remarks: @Wiley gave a very nice proof. Here we give an alternative proof.

Using the well-known inequality $a^2b + b^2c + c^2a \le \frac{4}{27}(a + b + c)^3 - abc$, it suffices to prove that
$$\frac{4}{27}(a + b + c)^3 - abc \le 3. \tag{1}$$
The condition $a^a b^b c^c = 1$ is written as $$a\ln a + b\ln b + c\ln c = 0. \tag{2}$$
Note that $x\mapsto x\ln x$ is convex on $x > 0$. By Jensen's inequality and (2), we have
$$0 = a\ln a + b\ln b + c\ln c
\ge 3 \cdot \frac{a + b + c}{3}\ln \frac{a + b + c}{3}$$
which results in
$$a + b + c \le 3. \tag{3}$$
Fact 1: It holds that
$x\ln x \ge x - 1 + \frac14(x - 1)^2$ for all $x \in (0, 3]$.
(The proof is given at the end.)
Using Fact 1 and (2) and (3), we have
$$
 0 = a\ln a + b\ln b + c\ln c \ge a + b + c - 3 + \frac14(a-1)^2 + \frac14(b-1)^2 + \frac14(c-1)^2
$$
or
$$a^2 + b^2 + c^2 + 2a + 2b + 2c - 9 \le 0. \tag{4}$$
We use the pqr method.
Let $p = a + b + c, q = ab + bc + ca, r = abc$.
We have
\begin{align*}
 &\frac{4}{27}(a+b+c)^3 - abc - 3\\
 =\,& \frac{4}{27}p^3 - r - 3\\
 \le\,& \frac{4}{27}p^3 - \frac{4pq - p^3}{9} - 3 \tag{5}\\
 =\,& \frac{7}{27}p^3 - \frac49 pq - 3\\
 \le\,& \frac{7}{27}p^3 - \frac49 p \cdot \frac{p^2 + 2p - 9}{2} - 3 \tag{6}\\
 =\,& \frac{1}{27}(p - 3)(p^2 - 9p + 27)\\
 \le\,& 0. \tag{7}
\end{align*}
Explanations:
(5): Degree three Schur inequality yields $r \ge \frac{4pq - p^3}{9}$.
(6): From (4), we have
$p^2 - 2q + 2p - 9\le 0$
which results in
$q \ge \frac{p^2 + 2p - 9}{2}$.
(7): From (3), we have $p \le 3$.
We are done.

Proof of Fact 1:
It suffices to prove that
$$\ln x \ge \frac{1}{x}\left(x - 1 + \frac14(x - 1)^2\right).$$
Let $F(x) := \mathrm{LHS} - \mathrm{RHS}$. We have
$$F'(x) = \frac{(x-1)(3-x)}{4x^2}.$$
We have $F'(x) < 0$ on $(0, 1)$,
and $F'(x) > 0$ on $(1, 3)$,
and $F'(1) = 0$, and $F'(3) = 0$.
Thus, $F(x) \ge F(1) = 0$ for all $x\in (0, 3]$.
We are done.
