How to prove four variable inequality involving sums of cube roots

Suppose that $a,b,c,d>0$. Is there a proof that $$ a\sqrt[3]{\frac{1+d}{b^3+abcd}}+b\sqrt[3]{\frac{1+d}{c^3+abcd}}+c\sqrt[3]{\frac{1+d}{a^3+abcd}}\geq 3?$$ I tried for example Jensen, Karamata, Power mean and Minkowski's inequality without success.

The inequality can be written as $$(1+d)\left(\sum_{\text{cyc}}\frac{a}{\sqrt[3]{b^3+abcd}}\right)^3\geq27$$ By the generalized Hölder's inequality $$\left(\sum_{\text{cyc}}\frac{a}{\sqrt[3]{b(b^2+d\,ca)}}\right)^3\left(\sum_{\text{cyc}}ab\right)\left(\sum_{\text{cyc}}a(b^2+d\,ca)\right)\geq\left(\sum_{\text{cyc}}a\right)^5$$ This reduces the problem to $$(1+d)\left(\sum_{\text{cyc}}a\right)^5\geq27\left(\sum_{\text{cyc}}ab\right)\left(\sum_{\text{cyc}}a(b^2+d\,ca)\right)$$ But $$\sum_{\text{cyc}}a(b^2+d\,ca)=\sum_{\text{cyc}}ab^2+d\sum_{\text{cyc}}ca^2=(1+d)\sum_{\text{cyc}}ab^2$$ So it's just $$\left(\sum_{\text{cyc}}a\right)^5\geq27\left(\sum_{\text{cyc}}ab\right)\left(\sum_{\text{cyc}}ab^2\right)$$

And see Inequality problem $(a+b+c)^5\geq27(ab+bc+ca)(ab^2+bc^2+ca^2)$.