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

We employ of the rearrangement inequality.
First, since $x\mapsto x^2$ preserves the order for $x>0,$ we have $$a^2b+b^2c+c^2a\le a^3+b^3+c^3.$$
Next write the condition as $$a\ln a+b\ln b+c\ln c=0.$$ Again using the rearrangement inequality, as $x\mapsto\ln x$ preserves the order for $x>0,$ we have the inequalities: $$\begin{cases}a\ln b+b\ln c+c\ln a\le0\\
a\ln c+b\ln a+c\ln b\le0\\
a\ln a+b\ln b+c\ln c=0\end{cases}$$
Adding these together, we have $(a+b+c)\ln(abc)\le0,$ hence $abc\le1.$
Now let $\ln a+\ln b+\ln c=k\le0.$ Apply the Lagrange multiplier method with condition $g(a,b,c):=\ln a+\ln b+\ln c=k$ for a fixed $k\le0,$ to maximize $f(a,b,c):=a^3+b^3+c^3.$
Thus the Lagrange multiplier gives that the extreme of $f$ occurs when $$\begin{cases}3a^2-\lambda\frac{1}{a}=0\\
3b^2-\lambda\frac{1}{b}=0\\
3c^2-\lambda\frac{1}{c}=0\end{cases},$$
i.e. when $a^3=b^3=c^3=\lambda;$ then $a=b=c$ and hence $f(a,b,c)=3abc=3e^k\le3.$
$\square$  
Hope this helps.  
Edit:
As pointed out in the comment, the final part about Lagrange multipliers is incorrect; in fact, given the constraint $abc<1,$ it does not follow that $a^2b+b^2c+c^2a\le3.$
Also pointed out in the comment is to use Jensen's inequality to obtain $a+b+c\le3,$ but we are not getting answers yet.
Everything we tried so far to fix this fails. We shall update if we find a way to work around.
