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
 A: Rozenberg's inequality is solved, find the solution at this link :
If $a^3+b^3+c^3=3$ so $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}$
A: Alternative proof: the pqr method
Let $p = a + b + c$ and $q = ab + bc + ca$ and $r = abc$.
We will use $p^2 \ge 3q$ and $q^2 \ge 3pr$ which are well-known. See the remarks at the end.
Using $a^3 + b^3 + c^3 = p^3 - 3pq + 3r$, the condition is written as
$$p^3 - 3pq + 3r = 3. \tag{1}$$
Using (1) and $q^2 \ge 3pr$, we have
$$p^3 - 3pq + 3\cdot \frac{q^2}{3p} \ge 3$$
or
$$\frac{1}{p}\left(\frac32p^2 - q\right)^2 \ge \frac54 p^3 + 3$$
which results in (using $p^2 \ge 3q$)
$$\frac32p^2 - q \ge \frac12\sqrt{5p^4 + 12p}$$
or
$$q \le \frac{3}{2}p^2 - \frac12\sqrt{5p^4 + 12p}. \tag{2}$$
Using (2), we have
\begin{align*}
 &2(a^2 + b^2 + c^2) - (ab + bc + ca) - 3\\
 =\,& 2(p^2 - 2q) - q - 3\\
 =\,& 2p^2 - 5q - 3\\
 \ge\,& 2p^2 - 5\cdot \left(\frac{3}{2}p^2 - \frac12\sqrt{5p^4 + 12p}\right) - 3 \\
 =\,& \frac52\sqrt{5p^4 + 12p} - \frac{11}{2}p^2 - 3\\
 \ge\,& 0
\end{align*}
where we have used
$$\left(\frac52\sqrt{5p^4 + 12p} \right)^2 - \left( \frac{11}{2}p^2 + 3\right)^2
= (3 - p)(-p^3 - 3p^2 + 24p - 3) \ge 0.$$
(Note: It is easy to prove that $1 \le p \le 3$. Also,
$-p^3 - 3p^2 + 24p - 3 \ge 0$ for all $1 \le p \le 3$. See the remarks at the end.)
We are done.

Remarks:

*

*$p^2 \ge 3q$ is just $a^2 + b^2 + c^2 \ge  ab + bc + ca$.


*$q^2 \ge 3pr$ is just $(ab)^2 + (bc)^2 + (ca)^2 \ge ab \cdot bc + bc \cdot ca + ca \cdot ab$.


*We have $p \le 3$ which follows from the power mean inequality $\sqrt[3]{\frac{a^3 + b^3 + c^3}{3}} \ge \frac{a + b + c}{3}$.


*We have $p \ge 1$.
Indeed, if $p < 1$, then $a, b, c \le 1$ and
$1 > a + b + c \ge a^3 + b^3 + c^3$ which contradicts the condition.
