For all positive reals $a, b, c$, we wish to prove the inequality
$$\sum_{\text{cyc}} \sqrt{\frac{a^3}{b^3 + (c+a)^3}} \ge 1$$
My approach was Hölder:
$$\left(\sum_{\text{cyc}} \sqrt{\frac{a^3}{b^3 + (c+a)^3}}\right)^2 \cdot \left(\sum_{\text{cyc}} b^3 + (c+a)^3 \right)^1 \ge (a+b+c)^3$$
So it would suffice to prove that $(a + b + c)^3 \ge a^3 + b^3 + c^3 + \sum_{\text{cyc}}(a+b)^3$. This is unfortunately not true. Another approach I came up with was by Titu's lemma:
$$\sum_{\text{cyc}} \sqrt{\frac{a^3}{b^3 + (c+a)^3}} = \sum_{\text{cyc}} \frac{a^2}{\sqrt{a(b^3 + (c+a)^3)}} \ge \frac{(a+b+c)^2}{\sum_{\text{cyc}} \sqrt{a(b^3 + (c+a)^3)}}$$
So it would suffice to prove $(a+b+c)^2 \ge \sum_{\text{cyc}} \sqrt{a(b^3 + (c+a)^3)}$. This is, however not true. Take, for example, $a, b, c = 1, 2, 3$ to get $LHS = 36, RHS = 38.32$. The problem I find is that the inequality seems to be pretty sharp, since both my attempts with Hölder and C-S overshoot in the "first step". To counter this, I thought of using Jensen. Because of homogeneity, we may assume $a + b + c = 1$, and we have to prove $\sum_{\text{cyc}} \sqrt{\frac{a^3}{b^3 + (1-b)^3}} \ge 1$. Although I was not able to make progress from here because I couldn't isolate each term to only one variable, as one often wishes to do in a Jensen approach.