<s>[INST] If $a^3+b^3+c^3=3$ so $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}$ 
Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=3$. Prove that:
  $$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}.$$

This inequality we can prove by BW with computer.
I am looking for an human proof, which we can use during competition.
For example, we need to prove that
$$\sum_{cyc}\left(\frac{a^3}{a+b}-\frac{a^2}{2}\right)\geq\frac{3}{2}-\frac{a^2+b^2+c^2}{2}$$ or
$$\sum_{cyc}\frac{a^2(a-b)}{a+b}\geq\sqrt[3]{3(a^3+b^3+c^3)^2}-a^2-b^2-c^2$$ or
$$\sum_{cyc}\frac{a^2(a-b)}{a+b}\geq\frac{3(a^3+b^3+c^3)^2-(a^2+b^2+c^2)^3}{\sqrt[3]{9(a^3+b^3+c^3)^4}+(a^2+b^2+c^2)\sqrt[3]{3(a^3+b^3+c^3)^2}+(a^2+b^2+c^2)^2}$$
and since $\sqrt[3]{3(a^3+b^3+c^3)^2}\geq a^2+b^2+c^2$ or $3(a^3+b^3+c^3)^2\geq(a^2+b^2+c^2)^3$,
it remains to prove that
$$\sum_{cyc}\frac{a^2(a-b)}{a+b}\geq\frac{3(a^3+b^3+c^3)^2-(a^2+b^2+c^2)^3}{3(a^2+b^2+c^2)^2}$$
and from here I don't see what to do. 
 [/INST] I propose a simple alternative of the third step of my proof in order to avoïd Lagrange multipliers. 
New Step 3. (version without Lagrange multipliers.) 
I suppose here that $\max (a,b,c)\geq \sqrt{2}.$ 
without loss of  generality I suppose that $a\geq \sqrt{2}.$ 
We also have  $a\leq 3^{1/3}$ and  $b,c \leq (3-\sqrt{2}^3)^{1/3}.$ 
Thus $\max(a+b,a+c,a+c)\leq  3^{1/3}+(3-\sqrt{2}^3)^{1/3}<2$
Hence 
$$a^3(a+b)+b^3(b+c)+c^3(a+c)< 2(a^3+b^3+c^3)=6.$$ 
c.q.f.d  
</s>