How prove this $(abc)^4+abc(a^3c^2+b^3a^2+c^3b^2)\le 4$ let $a,b,c>0$,and such that $a+b+c=3$,prove that
$$(abc)^4+abc(a^3c^2+b^3a^2+c^3b^2)\le 4$$
I first consider 
$$abc\le\left(\dfrac{a+b+c}{3}\right)^3=1$$
 so it suffices to show that
$$a^3c^2+b^3a^2+c^3b^2\le 3$$
 But I find this not true.
 A: Let $\{a,b,c\}=\left\{x,y,z\right\}$, where $x\geq y\geq z$.
Hence, by Rearrangement and AM-GM we obtain:
$$a^2c+b^2a+c^2b+abc=a\cdot ac+b\cdot ba+c\cdot cb+xyz\leq x\cdot xy+y\cdot xz+z\cdot yz+xyz=$$
$$=y(x+z)^2=4x\left(\frac{x+z}{2}\right)^2\leq4\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4.$$
Thus, since 
$$\sum_{cyc}ab\sum_{cyc}a^2c=\sum_{cyc}(a^3c^2+a^3bc+a^2b^2c),$$
we obtain:
$$a^4b^4c^4+abc\sum_{cyc}a^3c^2= a^4b^4c^4+abc\left(\sum_{cyc}ab\sum_{cyc}a^2c-\sum_{cyc}(a^3bc+a^2b^2c)\right)\leq$$
$$\leq a^4b^4c^4+abc\left((4-abc)\sum_{cyc}ab-\sum_{cyc}(a^3bc+a^2b^2c)\right)=$$
$$=a^4b^4c^4+abc\left(4(ab+ac+bc)-9abc\right).$$
Id est, it remains to prove that:
$$a^4b^4c^4+abc\left(4(ab+ac+bc)-9abc\right)\leq4.$$
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the last inequality is a linear inequality of $v^2$, 
which says that it's enough to prove this inequality for extremal value of $v^2$,
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$, where $0<a<\frac{3}{2}$.
Hence, we need to prove that
$$(a-1)^2\left(4+8a+12a^2-56a^3+41a^4+6a^5+7a^6+8a^7-72a^8+64a^9-16a^{10}\right)\geq0,$$
which is true for $0<a<\frac{3}{2}$.
Done!
