Prove $\sum\limits_{\mathrm{cyc}} \frac{4a+1}{\sqrt{4a+bc}}\ge 6$ for $a, b, c \ge 0$ with $ab+bc+ca=1$

algebra-precalculusholder-inequalityinequalityradicals

Problem. Let $a,b,c$ be non-negative real numbers satisfying $ab+bc+ca=1.$ Prove that
$$\frac{4a+1}{\sqrt{4a+bc}}+\frac{4b+1}{\sqrt{4b+ca}}+\frac{4c+1}{\sqrt{4c+ab}}\ge 6.$$

I've tried to use Holder inequality without success.

$$\left(\sum_{cyc}\frac{4a+1}{\sqrt{4a+bc}}\right)^2.\sum_{cyc}(4a+bc)(4a+1)\ge [4(a+b+c)+3]^3. \tag{1}$$

$$\left(\sum_{cyc}\frac{4a+1}{\sqrt{4a+bc}}\right)^2.\sum_{cyc}(4a+bc)(4a+1)(b+c)^3\ge \left(\sum_{cyc}(4a+1)(b+c)\right)^3.\tag{2} $$

Which is both leads to wrong inequality in general.

I'd like to ask two question.

  1. Is there a better Holder using ?

I think the appropriate one might be ugly but if you find it, please free share it here.

  1. Is there others idea which is smooth enough?

For example, Mixing variables, AM-GM or Cauchy-Schwarz…etc.

I aslo hope to see a good lower bound of $\dfrac{4a+1}{\sqrt{4a+bc}},$ which eliminate the radical form (may be the rest is simpler by $uvw$)

All idea and comment is welcome. Thank you for interest!

Remark. About $uvw$, see [here.][1]

Updated edit: The RiverLi's Holder using is impressed. If someone found other approach, please share it here.
[1]: https://artofproblemsolving.com/community/c6h278791

Best Answer

Some thoughts.

By Holder inequality, we have \begin{align*} &\left(\sum_{\mathrm{cyc}}\frac{4a + 1}{\sqrt{4a + bc}}\right)^2 \sum_{\mathrm{cyc}} (4a + 1)(4a + bc)(ab + 6bc + 3ca + a + 4b + 2c)^3\\ \ge{}& \left(\sum_{\mathrm{cyc}} (4a + 1)(ab + 6bc + 3ca + a + 4b + 2c)\right)^3. \tag{1} \end{align*}

It suffices to prove that \begin{align*} &\left(\sum_{\mathrm{cyc}} (4a + 1)(ab + 6bc + 3ca + a + 4b + 2c)\right)^3\\ \ge{}& 36\sum_{\mathrm{cyc}} (4a + 1)(4a + bc)(ab + 6bc + 3ca + a + 4b + 2c)^3. \tag{2} \end{align*}

(2) is true which is verified by Mathematica.

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