<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] Let $\{a,b,c\}=\{x,y,z\}$, where $x\geq y\geq z$.
Hence, by Rearrangement and AM-GM we obtain:
$$ab^2+bc^2+ca^2=ab\cdot b+bc\cdot c+ca\cdot a\leq xy\cdot x+xz\cdot y+yz\cdot z=y(x^2+xz+z^2)\leq$$
$$\leq y(x+z)^2=4y\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4.$$
Done!
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