For $a,b,c>0$ prove that $abc(a+b+c) \le a^3 b + b^3 c + c^3 a$ if $a,b,c &gt; 0$ then prove that $abc(a+b+c) \le a^3 b + b^3 c + c^3 a$

My attempt
The hint given for this question was to use the Cauchy-schwarz inequality. But if you look at the expression given, the L.H.S is symmetric but the R.H.S is not. If I understand the inequality(Cauchy-Schwarz)  correctly, both the sides of the inequality should be symmetric.
So can we use the Cauchy-Schwarz to prove this?
I also tried applying AM &gt; GM but got stuck.
Any help would be appreciated.
We can write this inequality as $$F=\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b} \ge (a+b+c)~~~(1)$$
Using Cauch-Schwarts ineq. : $$\frac{A^2}{X}+\frac{B^2}{Y}+\frac{C^2}{Z} \ge \frac{(A+B+C)^2}{X+Y+Z}, X,Y,Z&gt;0~~~~(2)$$
Using (2) it follows that $$F\ge \frac{(a+b+c)^2}{(a+b+c)}=a+b+c.$$
We use following inequality:
$a^4+b^4+c^4 ≥ abc(a+b+c)$
Let $a ≥b ≥c$, we may write:
$a^4 ≥a^3b$
$b^4 ≥b^3c$
$c^4&lt;c^3a$ ⇒ $c^4 ≥ -c^3a$
And we have:
$a^4+b^4+c^4+2c^3a ≥ a^3b+b^3c-c^3a+2c^3a =a^3b+b^3c+c^3a≥ abc(a+b+c)$