<s>[INST] Given $a+b+c =1$ prove $\displaystyle \sum_{\text{cyclic}}\sqrt{abc+4ab+4ac}\ge 8(ab+bc+ca)+\sqrt{abc}$ 
If $a,b,c$ are non-negative real numbers such that: $a+b+c=1.$ Prove that: $$\sqrt{abc+4ab+4ac}+\sqrt{abc+4bc+4ba}+\sqrt{abc+4ca+4cb}\ge8(ab+bc+ca)+\sqrt{abc}$$

Anyone can help me give a hint to get proof?
I guess equality holds iff: two of them are equal to $0.$ Since $ab+bc+ca\le\frac{1}{3}$ and $abc\le\frac{1}{27}$.
But the last inequality is not true by a countable example. Also, it is hard for me with this kind of equality. I tried squaring both side, the rest is very complicated.
Thank you for your interest!
 [/INST] Hint:
Observe,
$$\begin{align*}
8(ab+bc+ca)+\sqrt{abc}&=4a(b+c)+4b(c+a)+4c(a+b)+\sqrt{abc} \\
(b+c-a)^2&=(1-2a)^2=1-4a(1-a)=1-4a(b+c)
\end{align*}$$
Using the Cauchy-Schwarz inequality,
$$\sum_{\text{cyc}}\sqrt{\left((b+c-a)^{2}+4a(b+c)\right)\left(abc+4a(b+c)\right)}\geq \sum_{\text{cyc}} \left(\sqrt{abc}(b+c-a)+4a(b+c)\right)\\
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=8(bc+ca+ab)+\sqrt{abc}.$$
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