Let $a,b,c>0$, and $a+b+c=1$,show that $$\sum_{cyc}\dfrac{a^3+b^2}{b+c}\ge\dfrac{2}{3}+\dfrac{5+\sqrt{2}}{12}\sum_{cyc}(a-b)^2\tag{1}$$
I have prove $$\sum_{cyc}\dfrac{a^3+b^2}{b+c}\ge\dfrac{2}{3}$$ because Use Holder inequality
$$\sum_{cyc}\dfrac{a^3}{b+c}\cdot\sum(b+c)\sum_{cyc}(1)\ge (a+b+c)^3$$ so $$\sum_{cyc}\dfrac{a^3}{b+c}\ge\dfrac{1}{6}$$ and $$\sum_{cyc}\dfrac{b^2}{b+c}\ge\dfrac{(a+b+c)^2}{2(a+b+c)}=\dfrac{1}{2}$$ For $(1)$ I can't,
We'll prove a stronger inequality:
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=1$. Prove that $$\sum_{cyc}\frac{a^3+b^2}{b+c}\geq\frac{2}{3}+\frac{2}{3}\sum_{cyc}(a-b)^2.$$ Indeed, by C-S $$\sum_{cyc}\frac{b^2}{b+c}\geq\frac{(a+b+c)^2}{2(a+b+c}=\frac{1}{2}.$$ Thus, it's enough to prove that $$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{1}{6}+\frac{4}{3}\sum_{cyc}(a^2-bc)$$ or
$$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{(a+b+c)^2}{6}+\frac{4}{3}\sum_{cyc}(a^2-bc)$$ or $$\sum_{cyc}\frac{a^3}{b+c}\geq\frac{1}{2}\sum_{cyc}(3a^2-2ab)$$ or $$\sum_{cyc}\left(\frac{2a^3}{b+c}-3a^2+ab+ac\right)\geq0$$ or $$\sum_{cyc}\frac{a(2a^2+b^2+c^2-3ab-3ac+2bc)}{b+c}\geq0$$ or $$\sum_{cyc}\frac{a((c-a)(c+b-a)-(a-b)(b+c-a))}{b+c}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{b(a+c-b)}{a+c}-\frac{a(b+c-a)}{b+c}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2(a^2+b^2-c^2)}{(a+c)(b+c)}\geq0$$ or $$\sum_{cyc}(a-b)^2(a+b)(a^2+b^2-c^2)\geq0.$$ Now, let $a\geq b\geq c$.
Thus, $$\sum_{cyc}(a-b)^2(a+b)(a^2+b^2-c^2)\geq$$ $$\geq(a-c)^2(a+c)(a^2+c^2-b^2)+(b-c)^2(b+c)(b^2+c^2-a^2)\geq$$ $$\geq(b-c)^2(a+c)(a^2-b^2)+(b-c)^2(b+c)(b^2-a^2)=(b-c)^2(a-b)^2(a+b)\geq0.$$
