Inequality $\frac{1}{\sqrt{ab+2}}+\frac{1}{\sqrt{bc+2}}+\frac{1}{\sqrt{ca+2}}\ge \frac{2\sqrt{3}}{\sqrt{a+b+c+abc}}.$

Question

Let $a,b,c\ge 0: ab+bc+ca=3.$ Prove that $$\frac{1}{\sqrt{ab+2}}+\frac{1}{\sqrt{bc+2}}+\frac{1}{\sqrt{ca+2}}\ge \frac{2\sqrt{3}}{\sqrt{a+b+c+abc}}.$$

Equality occurs when $a=b=c=1.$ Also, there's an interesting fact that a big trouble is around $(0,\sqrt{3},\sqrt{3}).$

For these values, we obtain $LHS-RHS\approx 0.000217.$ I hope Holder with appropriate yield helps.

I've tried AM-GM $$\frac{1}{\sqrt{ab+2}}+\frac{1}{\sqrt{bc+2}}+\frac{1}{\sqrt{ca+2}}\ge 3\sqrt[6]{\frac{1}{(ab+2)(bc+2)(ca+2)}},$$ and Cauchy-Schwarz $$\frac{1}{\sqrt{ab+2}}+\frac{1}{\sqrt{bc+2}}+\frac{1}{\sqrt{ca+2}}\ge \frac{9}{\sqrt{ab+2}+\sqrt{bc+2}+\sqrt{ca+2}}$$which both leads to wrong inequality when $a=b=\sqrt{3};c=0.$

We can not use the easy result $$\frac{1}{\sqrt{ab+2}}+\frac{1}{\sqrt{bc+2}}+\frac{1}{\sqrt{ca+2}}\ge \sqrt{3}$$ since $a+b+c+abc\ge 4$ is wrong.

Hope to see some helps. Any ideas and comments are welcome and appreciate.

Answer 1

Bacteria helps again!

By Holder: $$\left(\sum_{cyc}\frac{1}{\sqrt{ab+2}}\right)^2\sum_{cyc}(ab+2)(c+3)^3\geq(a+b+c+9)^3.$$ Thus, it's enough to prove that: $$(a+b+c+9)^3(a+b+c+abc)\geq12\sum_{cyc}(ab+2)(c+3)^3.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$ and $abc=w^3$.

Thus, $v=1$ and we need to prove that:$$(3u+9v)^3(3uv^2+w^3)\geq12v\sum_{cyc}(ab+2v^2)(c+3v)^3,$$ which is a linear inequality of $w^3$.

Thus, it's enough to prove a last inequality for an extremal value of $w^3$, which by $uvw$

(see here: https://artofproblemsolving.com/community/c6h278791) happens in the following cases.

1. $w^3=0.$

Let $c=0$.

Thus, $ab=3$ and we need to prove $$\left(a+\frac{3}{a}+9\right)^3\left(a+\frac{3}{a}\right)\geq12\left(27\cdot5+2(a+3)^3+2\left(\frac{3}{a}+3\right)^3\right)$$ or $$a^8+3a^7+39a^6+324a^5-1404a^4+972a^3+351a^2+81a+81\geq0,$$ which is true and smooth.

2. Two variables are equal.

Let $b=a$.

Thus, $c=\frac{3-a^2}{2a},$ where $0<a\leq\sqrt3$ and we need to prove that: $$\left(2a+\frac{3-a^2}{2a}+9\right)^3\left(2a+\frac{3-a^2}{2a}+\frac{a(3-a^2)}{2}\right)\geq$$ $$\geq12\left((a^2+2)\left(\frac{3-a^2}{2a}+3\right)^3+2\left(\frac{3-a^2}{2}+2\right)(a+3)^3\right)$$ or $$(a-1)^2(3+12a+72a^2+444a^3+134a^4-60a^5-80a^6-12a^7-a^8)\geq0,$$ which is true because for $a=\sqrt3$ we have $$3+12a+72a^2+444a^3+134a^4-60a^5-80a^6-12a^7-a^8=48(10\sqrt3-17)>0$$ and we are done!