prove :$\frac{36}{a^3+b^3}+11(a+b)\ge24+4\sqrt{3ab}$

Let $a,b\ge0: a+b>0$. Prove that: $$\frac{36}{a^3+b^3}+11(a+b)\ge24+4\sqrt{3ab}$$ When does equality hold?

I am stuck in this problem. Due to non-homogeneous inequality, I can not guess where is equality case.

I put $s=a+b; p=ab$ then we need to prove: $$\frac{36}{s^3-3ps}+11s\ge24+4\sqrt{3p}$$ But how should I finish the rest? Thanks for your help!

Edit: add more solutions

Here is my friend's solutions.

I continue your process. Let $m=\sqrt{3p},m=ks$ then the inequality become$$\frac{36}{s^3(1-k^2)}+(11-4k)s\ge24$$and accroding to the condition,we have$$0\leq k\leq \frac{\sqrt{3}}{2},s>0$$then we use the AM–GM inequality$$\frac{36}{s^3(1-k^2)}+(11-4k)s/3+(11-4k)s/3+(11-4k)s/3\ge4 \sqrt[4]{\frac{36 (11-4 k)^3}{27 \left(1-k^2\right)}}$$for$$\frac{(11-4 k)^3}{1-k^2}$$It's not difficult to prove that it gets the minimum at $k=\frac{1}{2}$and so$$4 \sqrt[4]{\frac{36 (11-4 k)^3}{27 \left(1-k^2\right)}}\\\ge24$$so the inequality has been proved and when $$k=\frac{1}{2},s=2$$i.e$$\text{ab}=\frac{1}{3},a+b=2$$we obtain the equality case