Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$.

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 $$a^2b+b^2c+c^2a+abc\le b(a^2+c^2+ac)+abc=b(a+c)^2=2b(a+c)(a+c)/2\le\dfrac{4}{27}(a+b+c)^3$$

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.