If $a^3+b^3+c^3=3$ so $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}$

Assume $$ (H):\;\; a,b,c>0, \; a^3+b^3+c^3=3,$$ and prove that

$$(I_0): \dfrac{a^3}{a+b}+\dfrac{c^3}{a+c}+\dfrac{b^3}{b+c}\geq\dfrac{3}{2}$$

Proof Step 1

By Cauchy-Schwarz we can see that the following inequality $(I_1)$ $$ (I_1):\;\; s(a,b,c) =a^3 (a+b)+c^3 (a+c)+b^3 (b+c)\leq 6. $$ implies ($I_0$). Indeed we have $$9=(\sum_{cycl} a^3)^2 \leq \sum_{cycl} \dfrac{a^3}{a+b}\times \sum_{cycl} a^3(a+b).$$ Thus, I will show $(I_1)$

Step 2. Suppose $\max (a,b,c)< \sqrt{2}.$

Let $x\geq 0.$ Young ineg. allows to write:
$$ a^3 x \leq \dfrac{1}{2} a^3 +\dfrac{1}{2} a^3 x^2.$$ Applying again Young ineg. in the the previous ineq. (second term of the right hand side) of the previous ineq. it follows $$ a^3 x \leq \dfrac{1}{2} a^3 +\dfrac{1}{2}(\dfrac{1}{2} a^3+ {\dfrac{1}{2} a^3 x^4})= (\dfrac{1}{2}+ \dfrac{1}{2^2})a^3+ \dfrac{1}{2^2} a^3 x^4 ,$$

and by iteration
$$a^3 x \leq \sum_{j=1}^n \dfrac{1}{2^j} a^3+ \dfrac{1}{2^n} x^{2^n}=(1-2^{-n}) a^3 + \dfrac{1}{2^n}a^3 x^{2^n}.$$

I apply that to $s(a,b,c)$ and get: $$ s(a,b,c)\leq (1-2^{-n})\times 2(a^3+b^3+c^3) +2 \times 3 \times \dfrac{1}{2^n}(a^{2^n}+b^{2^n}+c^{2^n})\leq 6,$$ passing to the limit :

There is a mistake in my proof (step 2). There was a missprint ($2n$ instead of $2^n$) which is now corrected. Now the inequality is exact but ineffective.

Step 3. I suppsose 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 $b,c \leq (3-\sqrt{2}^3)^{1/3}.$

First we can easily see that if $b$ or $c$ vanishes then
$$s(a,b,c)<6.$$ It is left to the reader.

So I suppose that $b>0$ and $c>0.$

To finish the proof we will see with Lagrange multipliers method that the functional $s(a,b,c)$ has no critical point for (a,bc) in the domain of this step. Indeed let $$h(a,b,c)=s(a,b,c)-\lambda (a^3+b^3+c^3-3).$$

We need to solve $$\dfrac{\partial h}{\partial a}=a^3+3 a^2(a+b)+c^3-3 a^2 \lambda =0 $$ and $$\dfrac{\partial h} {\partial b}= a^3 + b^3 + 3 b^2 (b + c) - 3 b^2 \lambda. $$

Thus
$$ (Eq_0): \lambda=\dfrac{4 a}{3}+b+\dfrac{c^3}{3 a^2}=\dfrac{a^3}{3 b^2}+\dfrac{4 b}{3}+c.$$

But

$$\dfrac{4 a}{3}+b+\dfrac{c^3}{3 a^2}\leq \dfrac{4 \times3^{1/3}}{3}+(3-\sqrt{2}^3)^{1/3}+\dfrac{(3-\sqrt{2}^3)}{3 \sqrt{2}^2}\approx 2.50616$$ and $$\dfrac{a^3}{3 b^2}+\dfrac{4 b}{3}+c> \dfrac{a^3}{3 b^2}\geq \dfrac{1}{(3 - 2 \sqrt{2})^{2/3}}\approx 3.23868.$$ Hence, the identity $(Eq_0)$ is not possible. c.q.f.d

New Step 3. (New version without Lagrange multipliers.)

I suppsose here that $\max (a,b,c)\geq \sqrt{2}.$ suppsose here that $\max (a,b,c)\geq \sqrt{2}.$

without restriction to the 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

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