Find the maximal value of $c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a}$ in terms of $c^3 + a^3 = m$ where $a, b$ and $c$ are positives. 
Let $a, b$ and $c$ be positive real numbers such that $c^3 + a^3 = m$ and $$[(b^3 - abc) + ab(a + b) - c(a^2 + b^2)][(abc - b^3) - bc(b + c) + a(b^2 + c^2)] \le 0.$$ Find the maximal value of $$c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a}$$ in terms of $m$.

We have that $$[(b^3 - abc) + ab(a + b) - c(a^2 + b^2)][(abc - b^3) - bc(b + c) + a(b^2 + c^2)] \le 0$$
$$\iff (b^3 - c^3)(a^3 - b^3) \ge 0 \implies b^3(c^3 + a^3 - b^3) \ge (ca)^3 \iff m - b^3 \ge \left(\frac{ca}{b}\right)^3$$
It is evident that $$c^2 - ca + a^2 - \frac{(b + 1)^2(b - 2)}{c + a} = \frac{m - b^3 + 3b + 2}{c + a} \ge \frac{(ca)^3 + 3b^4 + 2b^3}{b^3(c + a)}$$
How should I continue?


Claim: The maximal value is $${\frak m}=\frac12\left(\frac m2\right)^{2/3}+\left(\frac m2\right)^{-1/3}+\frac32\tag1$$ attained at $a=b=c$.

Let $$d=c^2 - ca + a^2 - \dfrac{(b + 1)^2(b - 2)}{c + a}=\frac{m-b^3+3b+2}{a+c}\tag2$$ with $a^3+c^3=m$. Supposing that $$(b^3 - abc) + ab(a + b) - c(a^2 + b^2)=r\tag3$$$$(abc - b^3) - bc(b + c) + a(b^2 + c^2)=s\tag4$$ with $rs\le0$, subtracting $(3)$ from $(4)$ yields \begin{align}r-s&=-2(abc-b^3)+a^2b-ab^2-a^2c-b^2c+b^2c+bc^2-ab^2-ac^2\\&=2b^3+b(a-c)^2-ac(a+c)\tag5.\end{align} This forms the equality constraint for Lagrange multipliers; in particular, $${\cal L}(a,b,c,\lambda)=\frac{m-b^3+3b+2}{a+c}+\lambda(2b^3+b(a-c)^2-ac(a+c)-r+s)\tag6$$ with the partial derivatives $$\begin{bmatrix}\dfrac{\partial{\cal L}}{\partial a}\\\dfrac{\partial{\cal L}}{\partial b}\\\dfrac{\partial{\cal L}}{\partial c}\end{bmatrix}=\begin{bmatrix}-\dfrac{m-b^3+3b+2}{(a+c)^2}+\lambda(2b(a-c)-c(2a+c))\\\dfrac{-3b^2+3}{a+c}+\lambda(6b^2+(a-c)^2)\\-\dfrac{m-b^3+3b+2}{(a+c)^2}+\lambda(-2b(a-c)-a(a+2c))\end{bmatrix}\tag7$$ which give, respectively, $$m+b^3-3b-2+\lambda(a+c)^2(2ab-2ac-2bc-c^2)=0\tag8$$$$-3b^2+3+\lambda(a+c)(6b^2+(a-c)^2)=0\tag9$$$$m+b^3-3b-2+\lambda(a+c)^2(-2ab-2ac+2bc-a^2)=0\tag{10}.$$ Subtracting $(8)$ from $(10)$ yields $$\lambda(a+c)^2(4ab-4bc+a^2-c^2)=\lambda(a+c)^2(a-c)(4b+a+c)=0\tag{11}$$ and since $a,b,c>0$, either $\lambda=0$ or $a=c$. If the former, $(8)$ implies that $m+b^3-3b-2=0$, so $d=0$. This is clearly not maximal as $a=b=c=1$ yields $d=3>0$. Therefore $$a=c\implies a^3+a^3=m\implies a^3=\frac m2.\tag{12}$$ Replacing $c$ by $a$ in $(3)$ and $(4)$ gives $$r=b^3-a^2b+a^2b+ab^2-a^3-ab^2=b^3-a^3\tag{13}$$$$s=a^2b-b^3-a^2b-ab^2+ab^2+a^3=a^3-b^3\tag{12}$$ so that $$(b^3-a^3)(a^3-b^3)=-(b^3-a^3)^2\le0\tag{14}.$$ This forces equality to hold at $a=b$, and together with $(12)$, the maximal value is $${\frak m}=\frac{m-a^3+3a+2}{a+a}=\frac{m-\frac m2+3\left(\frac m2\right)^{1/3}+2}{2\left(\frac m2\right)^{1/3}}\tag{15}$$ which proves $(1)$ and hence the claim.
