Show that $\sum_{cyc} \sqrt{8a+b^3}\ge 9$

Question

Prove that $$\sqrt{8a+b^3}+\sqrt{8b+c^3}+\sqrt{8c+a^3}\ge 9 \text{ for } a,b,c\ge0 \text{ and } a+b+c=3.$$

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By Holder $$\left(\sum _{cyc}\sqrt{8a+b^3}\right)^2\left(\sum _{cyc}\frac{1}{8a+b^3}\right)\ge 27$$

Or $$\sum _{cyc}\frac{1}{8a+b^3}\le \frac{1}{3} \text{ WLOG } 0\le a\le b =a+u\le c=a+v\le 3$$

By full expanding it's obvious, because:

$$u^2-uv+v^2\ge 0$$

$$\cdots $$

$$72u^9-80u^8v-776u^7v^2-591u^6v^3+683u^5v^4+1403u^4v^5+1569u^3v^6+1168u^2v^7+424uv^8+72v^9\ge $$

$$\ge uv(v-u)(80u^6+\cdots 80v^6)\ge 0$$

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My solution is very ugly. Can i solve it by Holder but more beautiful than it does? Help me

Answer 1

By Holder $$\left(\sum_{cyc}\sqrt{8a+b^3}\right)^2\sum_{cyc}\frac{(2a^2+2b^2+3ab+4ac+bc)^3}{8a+b^3}\geq$$ $$\geq\left(\sum_{cyc}(2a^2+2b^2+3ab+4ac+bc)\right)^3=64(a+b+c)^6.$$ Thus, it's enough to prove that $$64(a+b+c)^3\geq27\sum_{cyc}\frac{(2a^2+2b^2+3ab+4ac+bc)^3}{8(a+b+c)^2a+9b^3},$$ which is obviously true by BW and computer.

It's interesting that Hoder with $(ka+mb+c)^3$ does not help.

I got these coefficients by the following way.

Holder for two sequences it's the following.

Let $a_i>0$, $b_i>$, $\alpha>0$ and $\beta>0$. Prove that: $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$

The equality occurs for $$(a_1,a_2,...,a_n)||(b_1,b_2,...,b_n).$$ Now, by Holder $$\left(\sum_{cyc}\sqrt{8a+b^3}\right)^2\sum_{cyc}\frac{(ma^2+nb^2+kab+lac+bc)^3}{8a+b^3}\geq$$ $$\geq\left(\sum_{cyc}(ma^2+nb^2+kab+lac+bc)\right)^3=\left(\sum_{cyc}((m+n)a^2+(k+l+1)ab)\right)^3.$$ The equality occurs when $$\left(\sqrt{8a+b^3},\sqrt{8b+c^3},\sqrt{8c+a^3}\right)||$$ $$||\left(\tfrac{(ma^2+nb^2+kab+lac+bc)^3}{8a+b^3},\tfrac{(mb^2+nc^2+kbc+lab+ca)^3}{8b+c^3},\tfrac{(mc^2+na^2+kca+lbc+ab)^3}{8c+a^3}\right),$$ which after substitution $$(a,b,c)=(1,2,0)$$ gives $$(4,4,1)||\left(\frac{(m+4n+2k)^3}{16},\frac{(4m+2l)^3}{16},(n+2)^3\right),$$ which gives $$m+4n+2k=4m+2l$$ and $$m+4n+2k=4(n+2).$$ Now, we need to find values of $m,$ $n$, $k$ and $l$ such that they are solutions of this system and the inequality

$$\left(\sum_{cyc}((m+n)a^2+(k+l+1)ab)\right)^3\geq81\sum_{cyc}\frac{(ma^2+nb^2+kab+lac+bc)^3}{8a+b^3}$$ is true.

For $$(m,n,k,l)=(2,2,3,4)$$ it's indeed true and we are done!