I have the following problem to solve :
Let $a,b,c>0$ with $a\geq b \geq c$ such that $a^3+b^3+c^3=3$ and $a^3\in[1,3-2p]$,$b^3\in[p,1]$,$c^3\in[p,1]$ then we have : $$a+b^{\frac{2}{3}}+c^{\frac{1}{3}}\geq 3$$Where $p$ verify :$$p^{\frac{1}{9}}+p^{\frac{2}{9}}+(3-2p)^{\frac{1}{3}}=3$$
The main idea is the following lemma :
Let $b,c$ verify the conditions above then we have : $$b^{\frac{2}{3}}+c^{\frac{1}{3}}\geq (\frac{b+c}{2})^{\frac{2}{3}}+(\frac{b+c}{2})^{\frac{1}{3}}$$
It's not hard to prove and with this lemma we can put $b=c$
Remains to study the following function :
$$f(p)=p^{\frac{1}{9}}+p^{\frac{2}{9}}+(3-2p)^{\frac{1}{3}}-3$$
After this I'm a little bit stuck .
If you have an other way or if you can complete my proof it would be nice
Thanks a lot for your time .