<s>[INST] A cyclic inequality $\sum\limits_{cyc}{\sqrt{3x+\frac{1}{y}}}\geqslant 6$ Given positive real numbers $x,y,z$ satisfying $x+y+z=3$, prove that $$ \sqrt{3x+\frac{1}{y}}+\sqrt{3y+\frac{1}{z}}+\sqrt{3z+\frac{1}{x}}\geqslant 6. $$
 [/INST] Remark: @chenbai gave a nice solution using the Buffalo Way (BW).
Here, I give an alternative solution using BW:
By Holder, we have
\begin{align}
\left(\sum_{\mathrm{cyc}} \sqrt{3x + 1/y}\right)^2 \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + 1/y}
\ge \left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3.
\end{align}
It suffices to prove that
$$\left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3 \ge 36 \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + 1/y}.$$
After homogenization, it suffices to prove that, for all $x, y, z > 0$,
$$\left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3 \ge 12(x+y+z) \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + \frac{(x+y+z)^2}{9y}}.$$
The Buffalo Way kills it. We are done.
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