<s>[INST] Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$. Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$. 
How to prove inequality
$$
ab^2+bc^2+ca^2\le 4.\tag{*}
$$

In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality
$$
27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**}
$$
$\color{gray}{\mbox{(Without using "universal" Lagrange multipliers method).}}$
Thanks!
 [/INST] I prove this stronger inequality,
With loss of out,let $a=\min{(a,b,c)}$
\begin{align*}&4(a+b+c)^3-27(a^2b+b^2c+c^2a+abc)\\
&=9a(a^2+b^2+c^2-ab-bc-ac)+(4b+c-5a)(a+b-2c)^2\ge 0
\end{align*}
other nice methods:
with  out loss of let $b=mid{(a,b,c)}$,then $(b-a)(b-c)\le 0$,so
$$a^2b+b^2c+c^2a+abc\le b(a^2+c^2+ac)+abc=b(a+c)^2=2b(a+c)(a+c)/2\le\dfrac{4}{27}(a+b+c)^3$$
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