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!

Let $a=3x, b=3y, c=3z$ and $f(x,y,z)=x^2y+y^2 z +z^2 x$
Without loss of generality let's assume that $x= \max { (x,y,z)}$
Then $f(x+\frac{z}{2}, y+\frac{z}{2}, 0) - f(x,y,z)=yz(x-y)+\frac{xz}{2}(x-z) + \frac{z^3}{8} + \frac{z^2y}{4} \ge 0$
And $f(x,y,0)= x^2y = 4 \frac{x}{2} \frac{x}{2} y \le 4\left( \frac{x+y}{3} \right)^3=\frac{4}{27}$
From AM-GM inequality.
