# Hard inequality :$\sum_{cyc}\frac{a}{\sqrt[3]{a+b}}\leq a+b^{\frac{2}{3}}+c$

I'm interested by the following problem :

Let $$a,b,c$$ be positive real numbers such that $$a+b+c=1$$ and $$a\geq b \geq c$$ then we have : $$\sum_{cyc}\frac{a}{\sqrt[3]{a+b}}\leq 1-b+b^{\frac{2}{3}}=a+b^{\frac{2}{3}}+c$$

I tried logarithmic majorization but it fails because $$a\leq \frac{a}{\sqrt[3]{a+b}}$$ and I can prove a weaker version with Jensen we have : $$\frac{a^{\frac{4}{3}}}{\sqrt[3]{a+b}}+\frac{b}{\sqrt[3]{b+c}}+\frac{c^{\frac{4}{3}}}{\sqrt[3]{c+a}}\leq 1-b+b^{\frac{2}{3}}=a+b^{\frac{2}{3}}+c$$ I tried also to be free from the condition but it gives nothing.

I have the beginning of a proof using weighted Karamata's inequality but it's partial and i'm a bit lost .

If you have a hints it would be appreciable .

Try $$(a,b,c)=(0.47,0.47,0.06).$$
In this case $$1-b+b^{\frac{2}{3}}-\sum_{cyc}\frac{a}{\sqrt[3]{a+b}}=-0.00020...$$