show this inequality $\sum_{cyc}\frac{a^3}{a^2+ab+b^2}\ge\sqrt{\sum a^3}$

Question

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

Answer 1

$uvw$ helps!

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $$\sum_{cyc}\frac{a^3}{a^2+ab+b^2}\geq\sqrt{\frac{a^3+b^3+c^3}{a+b+c}}$$ or $$\frac{243u^3v^4-162uv^6-81u^4w^3+27v^4w^3}{27(3u^2v^4-u^3w^3-v^6)}\geq\sqrt{\frac{9u^3-9uv^2+w^3}{u}}$$ or $$u(9u^3v^4-6uv^6-3u^4w^3+v^4w^3)^2\geq(3u^2v^4-v^6-u^3w^3)^2(9u^3-9uv^2+w^3)$$ or $f(w^3)\geq0,$ where $$f(w^3)=-u^6w^9+9u^7v^2w^6-2u^3v^6w^6+uv^8w^6-36u^6v^6w^3+27u^4v^8w^3-6u^2v^{10}w^3-v^{12}w^3+27u^5v^{10}-27u^3v^{12}+9uv^{14}.$$ Now, $$f'(w^3)=-3u^6w^6+18u^7v^2w^3-4u^3v^6w^3+2uv^8w^3-36u^6v^6+27u^4v^8-6u^2v^{10}-v^{12}\leq$$ $$\leq18u^7v^2w^3-3u^3v^6w^3-36u^6v^6+27u^4v^8-6u^2v^{10}=$$ $$=-3u^2v^2(12u^4v^4-9u^2v^6+2v^8-6u^5w^3+uv^4w^3).$$ We'll prove that $$12u^4v^4-9u^2v^6+2v^8-6u^5w^3+uv^4w^3\geq0.$$ Indeed, the expression $$12u^4v^4-9u^2v^6+2v^8-6u^5w^3+uv^4w^3$$ decreases as a function of $w^3$, which says that it's enough to prove the last inequality for a maximal value of $w^3$, which happens for equality case of two variables.

Since the last inequality is homogeneous, it's enough to assume $b=c=1$, which gives $$\frac{4(a+2)^4(2a+1)^2}{243}-\frac{(a+2)^2(2a+1)^3}{27}+\frac{2(2a+1)^4}{81}-\frac{2(a+2)^5a}{81}+\frac{(a+2)(2a+1)^2a}{27}\geq0$$ or $$(a-1)^2(5a^4+16a^3+33a^2+37a+17)\geq0,$$ which is obviously true.

Id est, $f$ decreases, which says again that it's enough to prove $$\sum_{cyc}\frac{a^3}{a^2+ab+b^2}\geq\sqrt{\frac{a^3+b^3+c^3}{a+b+c}}$$ for $b=c=1$, which gives $$(a-1)^2(3a^4+2a^3+3a^2+12a+7)\geq0$$ and we are done!

Answer 2

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!