Proving the inequality $4\ge a^2b+b^2c+c^2a+abc$ So, a,b,c are non-negative real numbers for which holds that $a+b+c=3$.
Prove the following inequality: $$4\ge a^2b+b^2c+c^2a+abc$$
For now I have only tried to write the inequality as $$4\left(\frac{a+b+c}3\right)^3\ge a^2b+b^2c+c^2a+abc$$ but I don't know what to do after that...

So we need to show
$$4(a+b+c)^3 \ge 27(a^2b+b^2c+c^2a+abc)$$
One way is to use the cyclic symmetry and WLOG assume $a$ to be the min of $a, b, c$.  Then we can write $b = a+x, c = a+y$, where $x, y \ge 0$.  Now the inequality reduces to 
$$9a(x^2-xy+y^2)+(x-2y)^2(4x+y) \ge 0$$
which is obvious.  Also from the above, we get that equality is possible iff $x=y=0$ or when $a=0, x=2y$,  i.e. when $(a, b, c)=(1, 1, 1)$ or is a permutation of $(0, 2, 1)$.
