Let $a, b, c > 0$ such that $\left( 4a + 5b \right) \left( 4b + 5c \right) \left( 4c + 5a \right) = 729$. Prove that $$abc \left( a^{2} + bc + ca \right) \left( b^{2} + ca + ab \right) \left( c^{2} + ab + bc \right) \le 27.$$
This question was closed recently due to lack of source, attempt etc. So I post it here.
After homogenization, it suffices to prove that, for all $a, b, c > 0$, $$abc \left( a^{2} + bc + ca \right) \left( b^{2} + ca + ab \right) \left( c^{2} + ab + bc \right) $$ $$\le 27 \left(\frac{\left( 4a + 5b \right) \left( 4b + 5c \right) \left( 4c + 5a \right) }{729}\right)^3.$$ We can prove it by the Buffalo Way (BW). In detail, letting $a = c + s, b = c + t$ for $s, t \ge 0$, the above inequality is equivalent to $f(c) \ge 0$ with non-negative coefficients (polynomials in $s, t$).
However, this is complicated. We want to find nice proofs.
For example, I remember the following problem: If $x, y, z > 0$, prove that $(2x^4 + 3y^4)(2y^4+3z^4)(2z^4+3x^4) \ge (3x+2y)(3y+2z)(3z+2x)(xyz)^3$. It suffices to prove that $2x^4 + 3y^4 \ge (3x + 2y)xy^2 $ (due to @Ivan Meir). Can we deal with OP's inequality similarly?