show this inequality $\sum_{cyc}\frac{a^3}{a^2+ab+b^2}\ge\sqrt{\sum a^3}$ let $a,b,c>0$.such $a+b+c=1$ show that
$$\sum_{cyc}\dfrac{a^3}{a^2+ab+b^2}\ge \sqrt{a^3+b^3+c^3}$$
I have show that not stronger inequality:
$$\sum\dfrac{a^3}{a^2+ab+b^2}=\sum_{cyc}\dfrac{a^4}{a^3+a^2b+ab^2}\ge \dfrac{(a^2+b^2+c^2)^2}{\sum (a^3+ab^2+a^2b)}=\dfrac{(a^2+b^2+c^2)^2}{(a+b+c)(a^2+b^2+c^2)}=\dfrac{a^2+b^2+c^2}{a+b+c}=a^2+b^2+c^2$$
But for $(1)$ I can't prove it
 A: Also, we can use SOS here.
Indeed, we need to prove that:
$$\sum_{cyc}\frac{a^3}{a^2+ab+b^2}-\frac{a^2+b^2+c^2}{a+b+c}\geq\sqrt{\frac{a^3+b^3+c^3}{a+b+c}}-\frac{a^2+b^2+c^2}{a+b+c}$$ or
$$\frac{(ab+ac+bc)\sum\limits_{cyc}(a^4b^2+a^4c^2-a^3b^2c-a^3c^2b)}{(a+b+c)\prod\limits_{cyc}(a^2+ab+b^2)}\geq$$
$$\geq\frac{(a^3+b^3+c^3)(a+b+c)-(a^2+b^2+c^2)^2}{(a+b+c)\left(\sqrt{(a^3+b^3+c^3)(a+b+c)}+a^2+b^2+c^2\right)}$$ or
$$\sum_{cyc}(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab}{\sqrt{(a^3+b^3+c^3)(a+b+c)}+a^2+b^2+c^2}\right)\geq0.$$
Now, by C-S $$\sqrt{(a^3+b^3+c^3)(a+b+c)}\geq a^2+b^2+c^2.$$
Thus, it's enough to prove that:
$$\sum_{cyc}(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab}{2(a^2+b^2+c^2)}\right)\geq0.$$ 
Now, let $a\geq b\geq c$.
Thus, $$\sum_{cyc}(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab}{2(a^2+b^2+c^2)}\right)\geq$$
$$\geq(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab}{2(a^2+b^2+c^2)}\right)+$$
$$+(a-c)^2\left(\frac{b^2(ab+ac+bc)}{(a^2+ab+b^2)(b^2+bc+c^2)}-\frac{ac}{2(a^2+b^2+c^2)}\right)\geq$$
$$\geq(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab}{2(a^2+b^2+c^2)}\right)+$$
$$+(a-b)^2\left(\frac{b^2(ab+ac+bc)}{(a^2+ab+b^2)(b^2+bc+c^2)}-\frac{ac}{2(a^2+b^2+c^2)}\right)\geq$$
$$\geq(a-b)^2\left(\frac{c^2(ab+ac+bc)}{(a^2+ac+c^2)(b^2+bc+c^2)}-\frac{ab+\frac{bc}{2}}{2(a^2+b^2+c^2)}\right)+$$
$$+(a-b)^2\left(\frac{b^2(ab+ac+bc)}{(a^2+ab+b^2)(b^2+bc+c^2)}-\frac{ac+\frac{bc}{2}}{2(a^2+b^2+c^2)}\right)=$$
$$=(a-b)^2(ab+ac+bc)\left(\tfrac{c^2}{(a^2+ac+c^2)(b^2+bc+c^2)}+\tfrac{b^2}{(a^2+ab+b^2)(b^2+bc+c^2)}-\tfrac{1}{2(a^2+b^2+c^2)}\right)\geq0.$$
and we are done!
