$a,b,c>0:a+b+c=3.$ Prove that: $\sum\sqrt{\frac{ab+2}{ab+c}}\ge \frac{3\sqrt{6}}{2}.$

Question

Problem. Let $a,b,c>0:a+b+c=3.$ Prove that: $$\sqrt{\frac{ab+2}{ab+c}}+\sqrt{\frac{bc+2}{bc+a}}+\sqrt{\frac{ca+2}{ca+b}}\ge \frac{3\sqrt{6}}{2}.$$

I tried a lot without success.

By Holder, $$\left(\sum_{cyc}\sqrt{\frac{ab+2}{ab+c}}\right)^2\sum_{cyc}(ab+c)(ab+2)^2\ge (ab+bc+ca+6)^3,$$ which leads to wrong inequality at $a=b=0.96$ $$\left(\sum_{cyc}\sqrt{\frac{ab+2}{ab+c}}\right)^2\sum_{cyc}(ab+c)(ab+2)^2(a+b)^3\ge (\sum_{cyc}(ab+2)(a+b))^3=(3q-3r+12)^3$$ is not good enough.

Also, by AM-GM $$\sum_{cyc}\frac{4(ab+2)}{3(ab+c)+2(ab+2)}=4\sum_{cyc}\frac{ab+2}{3c+5ab+4},$$which implies $$\sum_{cyc}\frac{ab+2}{3c+5ab+4}\ge \frac{3}{4}.$$The last inequality is not true when $a=b\rightarrow 1.5$

By homogenizing, we need to prove $$\sum_{cyc}\sqrt{\frac{13ab+4(ca+cb)+2(a^2+b^2+c^2)}{3ab+ca+cb+c^2}}\ge \frac{9\sqrt{2}}{2}$$ Is there an idea called "isolated fudging" to prove the OP? I really hope someone share it here. Thank you very much.

Answer 1

We use the isolated fudging.

It suffices to prove that, for all $a, b, c > 0$, $$\frac{2}{3\sqrt 6}\sqrt{\frac{ab + 2(a + b + c)^2/9}{ab + c(a + b + c)/3}}\ge \frac{12(a^2 + b^2) + 32ab + 35c(a + b)}{24(a^2 + b^2 + c^2) + 102(ab + bc + ca)}. \tag{1}$$ (Summing cyclically on (1), the desired result follows.)

(1) can be proved by Buffalo Way (BW).

Answer 2

I hope the following reasoning will help.

By Holder $$\sum_{cyc}\sqrt{\frac{ab+2}{ab+c}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{ab+2}{ab+c}}\right)^2\sum\limits_{cyc}\frac{(ab+2)^2(c+8)^3}{(ab+c)^2}}{\sum\limits_{cyc}\frac{(ab+2)^2(c+8)^3}{(ab+c)^2}}}\geq\sqrt{\frac{\left(\sum\limits_{cyc}\frac{(ab+2)(c+8)}{ab+c}\right)^3}{\sum\limits_{cyc}\frac{(ab+2)^2(c+8)^3}{(ab+c)^2}}}$$ and it's enough to prove that: $$2\left(\sum\limits_{cyc}\frac{(ab+2)(c+8)}{ab+c}\right)^3\geq27\sum\limits_{cyc}\frac{(ab+2)^2(c+8)^3}{(ab+c)^2},$$ which seems true.

It's true for $b=a$, $c=3-2a$, where $0<a<\frac{3}{2}.$

Also, it's true for $c\rightarrow0^+$ and $b=3-a$, where $0<a<3$.

I think, $uvw$ can kill it, but it's very complicated.