Proving $\sum_{cyc}\sqrt{\frac{a^3}{a^2+8b^2}}\ge\sqrt{\frac{a^3+b^3+c^3}{a^2+b^2+c^2}}$ for positive $a$, $b$, $c$

Question

Hard problem with inequalities:

Let $a,b,c>0.$ Prove that $$\sqrt{\dfrac{a^3}{a^2+8b^2}}+\sqrt{\dfrac{b^3}{b^2+8c^2}}+\sqrt{\dfrac{c^3}{c^2+8a^2}} \ge \sqrt{\dfrac{a^3+b^3+c^3}{a^2+b^2+c^2}}$$

I saw it's on AOPS here.

I just can prove the weaker problem, it's $$\sqrt{\dfrac{a^3}{a^2+8b^2}}+\sqrt{\dfrac{b^3}{b^2+8c^2}}+\sqrt{\dfrac{c^3}{c^2+8a^2}} \ge \sqrt{\dfrac{a+b+c}{3}}.$$ Assume that $a+b+c=3.$ And By AM-GM $$9a+a^2+8b^2 \ge 6\sqrt{a(a^2+8b^2)},$$ or $$\sqrt{\dfrac{a^3}{a^2+8b^2}} \ge \dfrac{6a^2}{9a+a^2+8b^2}.$$ So $$\sum\limits_{cyc}\sqrt{\dfrac{a^3}{a^2+8b^2}} \ge \sum\limits_{cyc}\dfrac{6a^2}{9a+a^2+8b^2}.$$ And by Cauchy-Schwarz, $$\sum\limits_{cyc}\dfrac{6a^2}{9a+a^2+8b^2} \ge \dfrac{6(a^2+b^2+c^2)^2}{a^4+b^4+c^4+9(a^3+b^3+c^3)+8(a^2b^2+b^2c^2+c^2a^2)} \ge 1,$$ which the last inequality is easy. I hope holder can kill it, but I can't find something :( This is the first time I have asked a problem. If there are something wrong, I will improve.

Answer 1

By Holder $$\sum\limits_{cyc}\sqrt{\tfrac{a^3}{a^2+8b^2}}=\sqrt{\tfrac{\left(\sum\limits_{cyc}\sqrt{\tfrac{a^3}{a^2+8b^2}}\right)^2\sum\limits_{cyc}(a^2+8b^2)(c^2+ca+2a^2)^3}{\sum\limits_{cyc}(a^2+8b^2)(c^2+ca+2a^2)^3}}\geq\sqrt{\tfrac{\left(\sum\limits_{cyc}(2a^3+a^2b+a^2c)\right)^3}{\sum\limits_{cyc}(a^2+8b^2)(c^2+ca+2a^2)^3}}$$ and it's enough to prove that $$(a^2+b^2+c^2)\left(\sum\limits_{cyc}(2a^3+a^2b+a^2c)\right)^3\geq(a^3+b^3+c^3)\sum\limits_{cyc}(a^2+8b^2)(c^2+ca+2a^2)^3$$ or $$\sum_{cyc}(12a^{10}b-39a^9b^2+38a^8b^3+28a^8c^3+42a^7b^4+30a^7c^4-8a^6b^5+41a^6c^5)+$$ $$+abc\sum_{cyc}(12a^8-57a^7b+15a^7c+42a^6b^2+54a^6c^2+57a^5b^3+57a^5c^3-36a^4b^4)+$$ $$+a^2b^2c^2\sum_{cyc}(-138a^5-9a^4b-48a^4c-108a^3b^2-180a^3c^2+60a^3bc+135a^2b^2c)\geq0,$$ which is true by BW.