<s>[INST] 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
 [/INST] The condition can be presented in form
$$a\ln a +b\ln b + c\ln c = 0.$$
Let
$$a\leq b\leq c,$$
then
$$\ln a \leq \ln b \leq \ln c,$$
and we can use Chebyshev sum inequality:
$$3(a\ln a +b\ln b + c\ln c) \geq (a+b+c)\ln abc,$$
$$abc\leq 1.$$
Evidently, the expression $a^2b+b^2c+c^2a$ achieves maximum when $abc=1.$
The standard way of solving the problem on a conditional extremum is the method of Lagrange multipliers, which reduces it to a system of equations.
The maximum value of function
$$f(a,b,c,λ)=a^2b+b^2c+c^2a+λ(abc-1)$$
for $a,b,c>0$ on the interval
is reached or at its edges, or in one of the points with zero partial derivatives
$$f'_a=0,\quad f'_b=0,\quad f'_c=0,\quad f'_λ=0,$$
or
$$\begin{cases}
2ab+c^2+\lambda bc = 0\\
2bc+a^2+\lambda ca = 0\\
2ca+b^2+\lambda ab = 0\\
abc-1 = 0,
\end{cases}$$
then
$$\begin{cases}
a(2ab+c^2) = b(2bc+a^2)=c(2ca+b^2),\\
abc=1,
\end{cases}$$
$$\begin{cases}
2b^2c -c^2a = a^2b\\
b^2c -2c^2a = -a^2b\\
abc=1,
\end{cases}$$
$$c^2a = b^2c = a^2b,\quad abc=1,$$
$$\dfrac cb =1, \dfrac ba =1, \quad abc=1,$$
$$a=b=c=1,$$
$$f(a,b,c,\lambda)=3.$$
Note that
$$\lim_{a\to 0} a\ln a = 0,$$
so on the edge $a=0$ we have to minimize
$f(b,c,\lambda) = b^2c + \lambda(bc-1),$
and then we get the system
$$\begin{cases}
2bc+\lambda c = 0\\
b^2+\lambda b = 0\\
bc-1=0
\end{cases}$$
with solution 
$$b=2,\quad c=\dfrac12,\quad f = 2 < 3.$$
That means that
$$\boxed{a^2b+b^2c+c^2a \leq 3}$$
</s>