Inequality : $ab^2+bc^2+cd^2+da^2 \leq 4$ I have a excercise like that: let $a;b;c \geq 0$ and $a+b+c=3$, prove: $ab^2+bc^2+ca^2 \leq 4$
We can assume $b=mid\{a;b;c\}$, then $(a-b)(b-c) \geq 0$, by that way, the problem is solved.
But I have a question: what should I do if the problem is for 4 variables, eg:

Let $a;b;c;d \geq 0$ and $a+b+c+d=3$, prove: $ab^2+bc^2+cd^2+da^2 \leq 4$

Thank you.

We have
If $x,\,y,\,z$ are non-negative real numers satisfy $x+y+z=3,$ then
$$xy^2+yz^2+zx^2\leqslant 4.$$
Now, suppose $b = \max \{a,\,b,\,c,\,d\}.$ From this inequality we put $x=a,\,y=b,\,z=c+d,$ we get
$$a^2(c+d)+ab^2+b(c+d)^2 \leqslant 4.$$
But
$$a^2(c+d)+ab^2+b(c+d)^2$$
$$= ab^2+bc^2+cd^2+da^2 + c(a^2+2bd)+d^2(b-c)$$
$$ \geqslant ab^2+bc^2+cd^2+da^2.$$
Therefore
$$ab^2+bc^2+cd^2+da^2 \leqslant 4.$$
The proof is completed.
