The desired inequality holds if (and only if) the following one does
$$2 \left( \frac{a^2 + b^2}{a+b} \right)^3 \geq a^3 + b^3 \quad (*)_{a,b}$$
for all $a, b > 0$. Indeed, $(*)_{a,b}$ is implied by the problem statement by setting $b = c$. Conversely, we recover the problem statement by summing $(*)_{a,b}$, $(*)_{b,c}$ and $(*)_{c,a}$. We now prove $(*)_{a,b}$ by observing that:
$$2(a^2 + b^2)^3 - (a^3 + b^3)(a+b)^3 = (a-b)^4(a^2 + ab + b^2) \geq 0.$$

One sees that the expressions in the statement are related by the following identity:
$$3(a^2 + b^2)(a + b) = 2(a^3 + b^3) + (a + b)^3$$
Combining this with the AM-GM inequality below
$$4 \left( \frac{a^2 + b^2}{a+b} \right)^3 + \frac{(a+b)^3}{2} + \frac{(a+b)^3}{2} \geq 3 \cdot \frac{a^2 + b^2}{a+b} \cdot (a+b)^2 = 3(a^2 + b^2)(a+b)$$
we recover the inequality $(*)_{a,b}$ as in the first solution.

Let $(a, b, c) = (1, 1, x)$ for $x > 0$. The inequality then becomes
$$2 \left( \frac{1+x^2}{1+x} \right)^3 \geq 1+x^3.$$
The key insight is that this inequality being true for all $x > 0$ is equivalent to the original inequality being true for all $a, b, c > 0$. Indeed, plugging in $x = a/b$, $x = b/c$, and $x = c/a$ respectively, we get
$$2 \left( \frac{a^2 + b^2}{a+b} \right)^3 \geq a^3 + b^3$$
$$2 \left( \frac{b^2 + c^2}{b+c} \right)^3 \geq b^3 + c^3$$
$$2 \left( \frac{c^2 + a^2}{c+a} \right)^3 \geq c^3 + a^3$$
Summing these three inequalities yields the desired original one. Now, we finish by
$$2(1 + x^2)^3 - (1 + x^3)(1 + x)^3 = (x - 1)^4(x^2 + x + 1) \geq 0.$$