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

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

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.

• Thanks. How did you find the yield $ab+6bc+3ca+a+4b+2c$ ? It seems unsual for a symmetrical inequality. Sep 16, 2023 at 13:27
• @TATAbox Try and error. I first tried $p_1a + p_2b + p_3c$ but failed. Then I tried $p_1a + p_2b + p_3 c + p_4ab + p_5bc + p_6ca$ and works. Sep 16, 2023 at 13:32
• @TATAbox Perhaps there are other forms? Sep 16, 2023 at 13:32
• I really hope see it. Btw, did you try the substitution $x=(4a+1)/\sqrt{4a+bc};...$ and find relatives between yields $x^2+y^2+z^2,x^2y^2z^2$ ? I believe it works in some way Sep 16, 2023 at 13:36
• @TATAbox I did not try. Sep 16, 2023 at 13:38