Problem. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=3.$ Prove that$$\color{black}{\frac{a}{\sqrt{8a+bc}}+\frac{b}{\sqrt{8b+ca}}+\frac{c}{\sqrt{8c+ab}}\le \frac{\sqrt{a+b+c+abc}}{2}.}$$
Source: Vo Quoc Ba Can.
This symmetrical inequality is extremally tight! It was in a test of Vo Quoc Ba Can's class. (Toan hoc muon mau).
Some thoughts.
A big trouble here is equality occuring $(a,b,c)=(1,1,1)$ or $(a,b,c)=(0,\sqrt{3},\sqrt{3})$ and its permutations.
I'm afraid of that the author has a nice proof as usual like AM-GM, C-S,…etc. Unfortunately, I don't see any bright ideas.
I think the Bacteria method helps for this situation well but I failed to find an appropriate using.
For example, we can trivially use Cauchy-Schwarz inequality as$$\color{black}{\sum_{cyc}\frac{a}{\sqrt{8a+bc}}=\sum_{cyc}\sqrt{\frac{a}{8a+bc}}\cdot\sqrt{a}\le \sqrt{(a+b+c)\sum_{cyc}\frac{a}{8a+bc}}\le \frac{\sqrt{a+b+c+abc}}{2}}$$which the last inequality saves the equality occuring but is wrong at $a=b=\dfrac{1}{2};c=\dfrac{11}{4}.$
Also, the equality cases $a=b=c; a=b;c=0$ recalls us about Schur inequality.
Indeed, by using Schur of third degree $$abc\ge \frac{(a+b+c)[4(ab+bc+ca)-(a+b+c)^2]}{9}=\frac{(a+b+c)[12-(a+b+c)^2]}{9},$$
it implies $$RHS\ge \frac{1}{6}\cdot\sqrt{(a+b+c)[21-(a+b+c)^2]}.$$
(I just mention in case $21-(a+b+c)^2\ge 0.$)
Now, I saw an interesting fact that $$\frac{a}{\sqrt{8a+bc}}+\frac{b}{\sqrt{8b+ca}}+\frac{c}{\sqrt{8c+ab}}\ge \frac{1}{6}\cdot\sqrt{(a+b+c)[21-(a+b+c)^2]}.$$
It means that the original inequality is more stronger than Schur$$RHS\ge LHS\ge \frac{1}{6}\cdot\sqrt{(a+b+c)[21-(a+b+c)^2]}.$$
How hard could it be!
I hope to see some ideas for this nice problem. Thank you for your interest.
Best Answer
Some thoughts.
By Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} &\left(\frac{a}{\sqrt{8a+bc}}+\frac{b}{\sqrt{8b+ca}}+\frac{c}{\sqrt{8c+ab}}\right)^2\\[6pt] \le{}& \left(\sum_{\mathrm{cyc}}\frac{a^2}{(8a + bc)(3a +abc )}\right)\left(\sum_{\mathrm{cyc}} (3a + abc)\right)\\[6pt] ={}& \left(\sum_{\mathrm{cyc}}\frac{a^2}{(8a + bc)(3a +abc )}\right)\cdot 3(a + b + c + abc). \end{align*}
It suffices to prove that $$\sum_{\mathrm{cyc}}\frac{a^2}{(8a + bc)(3a +abc )} \le \frac{1}{12}. \tag{1}$$
(1) is true which is verified by Mathematica.