Let $a,b\ge\dfrac{c}{2}\ge0: (c+a)(c+b)>0$. Prove that: $$\frac{1}{a^3+c^3}+\frac{1}{b^3+c^3}+2\left[a+b-c(1-\sqrt{3})\right]\ge3\left(\frac{1}{\sqrt[3]{a+c}}+\frac{1}{\sqrt[3]{b+c}}\right).$$ I didn't know when equality occurs. My tried is using Holder but it leads to nothing.
Hope to see some hints. Thank you.
On
We can comb this problem:
By AM-GM $$\frac{1}{a^3+c^3}+\frac{1}{b^3+c^3}+2\left[a+b-c(1-\sqrt{3})\right]\ge$$ $$\geq3\sqrt[3]{\left(\frac{1}{a^3+c^3}+\frac{1}{b^3+c^3}\right)\left(a+b+(\sqrt3-1)c\right)^2}$$ and it remans to prove that: $$\left(\frac{1}{a^3+c^3}+\frac{1}{b^3+c^3}\right)\left(a+b+(\sqrt3-1)c\right)^2\geq\left(\frac{1}{\sqrt[3]{a+c}}+\frac{1}{\sqrt[3]{b+c}}\right)^3.$$ Now, let $a=\frac{xc}{2}$ and $b=\frac{yc}{2}.$
Thus, $x\geq1$ and $y\geq1$ and we need to prove that: $$\left(\frac{1}{x^3+8}+\frac{1}{y^3+8}\right)\left(x+y-2+2\sqrt3\right)^2\geq\left(\frac{1}{\sqrt[3]{x+2}}+\frac{1}{\sqrt[3]{y+2}}\right)^3.$$ Now, by Holder $$\left(\frac{1}{x^3+8}+\frac{1}{y^3+8}\right)\left(x+y-2+2\sqrt3\right)^2\geq\left(\sqrt[3]{\frac{(x-1+\sqrt3)^2}{x^3+8}}+\sqrt[3]{\frac{(y-1+\sqrt3)^2}{y^3+8}}\right)^3$$ and it's enough to prove that: $$\frac{(x-1+\sqrt3)^2}{x^3+8}\geq\frac{1}{x+2},$$ which is just $x-1\geq0$ and we are done!
Proof.
By using AM-GM $$\frac{1}{a^3+c^3}+2.\sqrt{a^2-ac+c^2}\ge \frac{3}{\sqrt[3]{a+c}}.$$$$\frac{1}{b^3+c^3}+2.\sqrt{b^2-bc+c^2}\ge \frac{3}{\sqrt[3]{b+c}}.$$Thus, we just need to prove $$a+b+(\sqrt{3}-1)c\ge \sqrt{a^2-ac+c^2}+\sqrt{b^2-bc+c^2}.$$ It's enough to prove $$a+\frac{(\sqrt{3}-1)c}{2}\ge \sqrt{a^2-ac+c^2},$$which is equivalent to $$\frac{4-2\sqrt{3}}{4}c^2+\sqrt{3}ac\ge c^2,$$or $$\sqrt{3}c\left(a-\frac{c}{2}\right)\ge 0.$$Similarly, $$b+\frac{(\sqrt{3}-1)c}{2}\ge \sqrt{b^2-bc+c^2}.$$ Hence, the proof is done. Equality holds at $$ \begin{cases} a=b=\dfrac{c}{2}\\\dfrac{1}{a^3+c^3}=\sqrt{a^2-ac+c^2}\\\dfrac{1}{b^3+c^3}=\sqrt{b^2-bc+c^2} \end{cases} \iff \begin{cases} a=b=\sqrt[4]{\dfrac{\sqrt{3}}{27}}\\c=\sqrt[4]{\dfrac{16\sqrt{3}}{27}} \end{cases} $$