<s>[INST] If $a^3+b^3+c^3=3$ so $2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$ I was looking at this question, and I derived this inequality from that

Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=3.$ Prove that:
$$2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$$

I couldn't solve Rozenberg's inequality but I assumed it's true. So if  $a^3+b^3+c^3=3$, then
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}\tag {1}$$
and also by using this, if $b^3+a^3+c^3=3$, then
$$\frac{b^3}{b+a}+\frac{a^3}{a+c}+\frac{c^3}{c+b}\geq\frac{3}{2}\tag {2}$$
using $(1)$ and $(2)$, if  $a^3+b^3+c^3=3$, then
$$\frac{a^3+b^3}{a+b}+\frac{b^3+c^3}{b+c}+\frac{c^3+a^3}{c+a}\geq3$$
$$\Leftrightarrow \sum\limits_{cyc}(a^2-ab+b^2)\geq 3$$
$$\Leftrightarrow 2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$$
If I'm not making any silly mistake, this inequality should be true. I couldn't solve yet, but I wanted to share with you.
$\mathbf {EDIT:}$ To be clear, I think that I derived an inequality weaker than Rozenberg's inequality by using his inequality, and I presented it for those of you who might interest trying to solve inequalities by ownselves
 [/INST] Rozenberg's inequality is solved, find the solution at this link :
If $a^3+b^3+c^3=3$ so $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}$
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