show this $\prod(a^3+b^3+ab)^2\ge (\sum a^2b)^3( \sum ab^2)^3$ let $a,b,c>0$ show that :$$(a^3+b^3+abc)^2(b^3+c^3+abc)^2(c^3+a^3+abc)^2-(a^2b+b^2c+c^2a)^3\cdot(ab^2+bc^2+ca^2)^3\ge 0$$Maybe there's some kind of identity in there.so I use wolfampha Calculated the results and found no identity. and the results are very complex. Maybe this inequality has a good solution.such as this following identity
$$(x+y+z)(xy+yz+zx)=(x+y)(y+z)(z+x)+xyz$$

We need to prove that
$$\prod_{cyc}\left(\frac{a^2}{bc}+\frac{b^2}{ac}+1\right)^2\geq\left(\sum_{cyc}\frac{a}{b}\right)^3\left(\sum_{cyc}\frac{a}{c}\right)^3.$$
Let $\frac{a}{b}=x$, $\frac{b}{c}=y$ and $\frac{c}{a}=z$.
Hence, $xyz=1$ and we need to prove that
$$\prod_{cyc}\left(\frac{x}{z}+\frac{y}{x}+1\right)^2\geq(x+y+z)^3(xy+xz+yz)^3$$ or
$$\prod_{cyc}(x^2+yz+zx)^2\geq(x+y+z)^3(xy+xz+yz)^3.$$
Now, since by AM-GM
$$\sum_{cyc}x^4y^2=\frac{1}{6}\sum_{cyc}(4x^4y^2+y^4z^2+z^4x^2)\geq\sum_{cyc}x^3y^2z,$$ we obtain:
$$\prod_{cyc}(x^2+yz+zx)=\sum_{cyc}(x^4y^2+x^3y^3+x^4yz+2x^2y^2z+3x^3z^2y+x^2y^2z^2)\geq$$
$$\geq\sum_{cyc}(x^3y^3+x^4yz+3x^3y^2z+3x^3z^2y+x^2y^2z^2).$$
Thus, it's enough to prove that
$$\left(\sum_{cyc}(x^3y^3+x^4yz+3x^3y^2z+3x^3z^2y+x^2y^2z^2)\right)^2\geq(x+y+z)^3(xy+xz+yz)^3xyz.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, we need to prove that
$$(27v^6-27uv^2w^3+3w^6+27u^3w^3-27uv^2w^3+3w^6+27uv^2w^3-9w^6+3w^6)^2\geq$$
$$\geq729u^3v^6w^3$$ or $f(u)\geq0,$ where
$$f(u)=(v^6-uv^2w^3+u^3w^3)^2-u^3v^6w^3.$$
But $$f'(u)=2(v^6-uv^2w^3+u^3w^3)w^3(3u^2-v^2)-3u^2v^6w^3=$$
$$=2uw^6(u^2-v^2)(3u^2-v^2)+v^6w^3(3u^2-2v^2)\geq0,$$
which says that $f$ increases.
Thus, it's enough to prove $f(u)\geq0$ for a minimal value of $u$, 
which happens for equality case of two variables.
Since the last inequality is symmetric and homogeneous, we can assume $y=z=1$,
which gives
$$(x^4+8x^3+9x^2+8x+1)^2\geq(2x^2+5x+2)^3x.$$
Let $x^2+1=2tx$.
Hence, by AM-GM $t\geq1$ and we need to prove that
$$(4t^2+16t+7)^2\geq(4t+5)^3$$ or
$$(t-1)(4t^3+20t^2+38t+19)\geq0.$$
Done!
