How to prove $\frac{1}{\sqrt{3a^2+13b}}+\frac{1}{\sqrt{3b^2+13c}}+\frac{1}{\sqrt{3c^2+13a}} \ge \frac{3}{4}$?

Question

Let $a,b,c\ge 0: ab+bc+ca>0$ and $a+b+c=3.$ Prove that $$\frac{1}{\sqrt{3a^2+13b}}+\frac{1}{\sqrt{3b^2+13c}}+\frac{1}{\sqrt{3c^2+13a}} \ge \frac{3}{4}. \tag{*}$$ Here is what I've done so far.

By using Cauchy-Schwarz inequality $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge \dfrac{9}{x+y+z}$ we get $$LHS_{(*)}\ge \frac{9}{\sqrt{3a^2+13b}+\sqrt{3b^2+13c}+\sqrt{3c^2+13a}}$$ and we need to prove $\sqrt{3a^2+13b}+\sqrt{3b^2+13c}+\sqrt{3c^2+13a}\le 12.$

Now I try to homogenize $a+b+c$ and it turns out $$\sqrt{3a^2+\frac{13}{3}(a+b+c)b}+\sqrt{3b^2+\frac{13}{3}(a+b+c)c}+\sqrt{3c^2+\frac{13}{3}(a+b+c)a}\le 4(a+b+c).$$ By squaring both side, we'll prove $$3(a^2+b^2+c^2)+2\sum_{cyc}\sqrt{3a^2+\frac{13}{3}(a+b+c)b}\cdot\sqrt{3b^2+13\frac{13}{3}(a+b+c)c}\le \frac{35}{3}(a+b+c)^2.$$

I stopped here. Hope you help me prove it. Thank you.

Answer 1

The inequality $\sum\limits_{cyc}\sqrt{3a^2+13b}\leq12$ is wrong.

Try $c=0$ and $a=b=\frac{3}{2}.$

Bacteria helps!

By Holder: $$\left(\sum_{cyc}\frac{1}{\sqrt{3x^2+13y}}\right)^2\sum_{cyc}(3a^2+13b)(6a+4b+9c)^3\geq\left(\sum_{cyc}(4a+6b+9c)\right)^3$$ and it's enough to prove that: $$6859(a+b+c)^3\geq\frac{9}{16}\sum_{cyc}(3a^2+13b)(6a+4b+9c)^3,$$ which is true, but my proof of the last statement is not so nice.

Answer 2

Some thoughts.

We may use the so-called isolated fudging.

It suffices to prove that $$\frac{1}{\sqrt{3a^2 + 13b}} \ge {\frac {768\,{a}^{2}+3715\,ab+6124\,ac+912\,{b}^{2}+2161\,bc+3120\,{c} ^{2}}{6400\,{a}^{2}+16000\,ab+16000\,ac+6400\,{b}^{2}+16000\,bc+6400\, {c}^{2}}} . \tag{1}$$ (Note: Taking a cyclic sum on (1), we get the desired inequality.)

We can prove (1) by the Buffalo Way (BW).