Prove $\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+c^3+abc}+\frac{1}{b^3+c^3+abc} \leq \frac{1}{abc}$

I have to apologize in advance that the question is missing some vital information (like, $a,b,c$ are positive? natural? rational?) I took it off some video off youtube.

In the video, an 11 year old was in the process of solving this question: prove $\frac{1}{a^3+b^3+abc}+\frac{1}{a^3+c^3+abc}+\frac{1}{b^3+c^3+abc} \leq \frac{1}{abc}$.

I don't really see how we can solve it in a simple way without using differential multivariable calculus. And even with calculus - it isn't so simple and not very pleasant. If anyone knows the question in its full form and knows how to answer it simply, I will be very happy to see it.

In the video, the child said "wlog $abc=1$" and defined $x=a^3$, $y=b^3$, $z=c^3$ and so he had to prove that

$\frac{1}{x+y+1}+\frac{1}{x+z+1}+\frac{1}{y+z+1} \leq 1$

The $abc=1$ part seems to be not so general, but if that helps you solve the question in any simple way - have at it.

We have $$\sum \frac{1}{a^3+b^3+abc} \le \sum \frac{1}{a^2b+ab^2+abc} = \frac{1}{a+b+c} \sum \frac{1}{ab} = \frac{1}{abc}$$

Note that $$a^3+b^3 \ge a^2b+ab^2 \iff (a-b)^2(a+b) \ge 0$$