Prove that: $ab\sqrt{ab}+bc\sqrt{bc}+ca\sqrt{ca}\le abc+\frac{1}{2}\sqrt[3]{\frac{(a^{2}+bc)^{2}(b^{2}+ca)^{2}(c^{2}+ab)^{2}}{abc}}$ 
Let $a,b,c>0$. Prove that: $$a b \sqrt{a b}+b c \sqrt{b c}+c a \sqrt{c a} \leqslant a b c+\frac{1}{2} \sqrt[3]{\frac{\left(a^{2}+b c\right)^{2}\left(b^{2}+c a\right)^{2}\left(c^{2}+a b\right)^{2}}{a b c}}$$

I really don't have many ideas in this problem. First I thought of using AM-GM: $a^2+bc\ge 2a\sqrt{bc}$ and so on, or using Holder $(\dfrac{a^3}{bc}+2a+\dfrac{bc}{a})(\dfrac{b^3}{ac}+2b+\dfrac{ca}{b})(\dfrac{c^3}{ab}+2c+\dfrac{ab}{c })$ but all make the inequality into the form $x^3+y^3+z^3\le3xyz$, which is of course incorrect. I thought about using derivatives but they don't work either
Can anyone give me a different way of thinking?
 A: Using AM-GM, it suffices to prove that
$$ab\frac{a + b}{2} + bc\frac{b + c}{2} + ca\frac{c + a}{2}
\le abc + \frac12\sqrt[3]{\frac{(a^2 + bc)^2(b^2 + ca)^2(c^2 + ab)^2}{abc}}$$
or
$$a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2 - 2abc
\le \sqrt[3]{\frac{(a^2 + bc)^2(b^2 + ca)^2(c^2 + ab)^2}{abc}}.$$
Since the inequality is homogeneous, WLOG, assume that $abc = 1$.
Let $p = a + b + c, q = ab + bc + ca, r = abc = 1$.
We have $p, q \ge 3$.
The inequality is written as
$$pq - 5r \le \sqrt[3]{\frac{(p^3r + q^3 - 6pqr + 8r^2)^2}{r}}$$
or
$$pq - 5 \le \sqrt[3]{(p^3 + q^3 - 6pq + 8)^2}.$$
Using AM-GM, we have
$$p^3 + q^3 = (p + q)(p^2 - pq + q^2)
\ge (p + q)pq \ge 2\sqrt{pq}\, pq.$$
Using $p, q \ge 3$, we have
$$2\sqrt{pq}\, pq \ge 6pq.$$
It suffices to prove that
$$pq - 5 \le \sqrt[3]{(2\sqrt{pq}\,pq - 6pq + 8)^2}.$$
Letting $x = \sqrt{pq} \ge 3$, the inequality becomes
$$x^2 - 5 \le \sqrt[3]{(2x^3 - 6x^2 + 8)^2}$$
or
$$(x^2 - 5)^3 \le (2x^3 - 6x^2 + 8)^2$$
or
$$(3x^4 - 6x^3 - 12x^2 + 14x + 21)(x - 3)^2 \ge 0$$
which is true.
We are done.
