Problem. Let $a,b,c\ge 0: a+b+c+abc=4$. Prove that: $$\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{b+c}{a+bc}}+\sqrt{\frac{c+a}{b+ca}}\ge 3.$$
I think the problem is nice and very hard. Equality holds iff $a=b=c=1; (0,2,2).$
The idea of using AM-GM directly doesn't work, because $$(a+b)(b+c)(c+a)\ge (a+bc)(b+ca)(c+ab),$$is wrong when $a=b=\dfrac{3}{2}.$
Also: $$\sum_{cyc}\sqrt{\frac{a+b}{c+ab}}\ge \sum_{cyc}\frac{2(a+b)}{a+b+c+ab},$$leads to wrong inequality.
Is Holder inequality is a good way in this case ?
I tried $$\left(\sum_{cyc}\sqrt{\frac{a+b}{c+ab}}\right)^2.\sum_{cyc}(a+b)^2(c+ab)\ge 8(a+b+c)^3,$$which is not good enough.
I think Mixing variables is a good approach but I am not good at using this method.
A fact: $$\sqrt[3]{\frac{a+b}{c+ab}}+\sqrt[3]{\frac{b+c}{a+bc}}+\sqrt[3]{\frac{c+a}{b+ca}}\ge 3,$$is already wrong when $a=b=1.874$.
Hope to see more ideas. Thank you.
Best Answer
Some thoughts.
By Holder, we have $$\left(\sum_{\mathrm{cyc}}\sqrt{\frac{a+b}{c+ab}}\right)^2 \sum_{\mathrm{cyc}} (a + b)^2(c + ab)(c + 2)^3 \ge{} \left(\sum_{\mathrm{cyc}} (a + b)(c + 2)\right)^3.$$
It suffices to prove that $$\left(\sum_{\mathrm{cyc}} (a + b)(c + 2)\right)^3 \ge 9 \sum_{\mathrm{cyc}} (a + b)^2(c + ab)(c + 2)^3. \tag{1} $$ This inequality is true which is verified by Mathematica. It can be proved by the pqr method.
Let $p = a + b + c, q = ab + bc + ca, r = abc$.
The condition $a + b + c + abc = 4$ becomes $p + r = 4$.
Using $r = 4- p$, (1) is written as \begin{align*} &208\,{q}^{3}+ \left( -72\,{p}^{2}+96\,p-2304 \right) {q}^{2}+ \left( 480\,{p}^{2}-5616\,p+15552 \right) q\\ &\quad +2384\,{p}^{3}-14184\,{p}^{2}+ 34560\,p-31104 \ge 0. \end{align*}
Omitted.