Problem. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=2.$ Find the minimum $$P=\frac{a}{a+ab+2}+\frac{b}{b+bc+2}+\frac{c}{c+ca+2}.$$
By $a=2,b=1,c=0$ I got a value $\dfrac{2}{3}.$
I've tried to prove $$\frac{a}{a+ab+2}+\frac{b}{b+bc+2}+\frac{c}{c+ca+2}\ge \frac{2}{3},$$which has full expanding expression is$$3[11abc+2abc(a+b+c)+8+4(a+b+c)+2(a^2c+c^2b+b^2a)]$$$$\ge 2[(abc)^2+9abc+3abc(c+a+b)+2(bc^2+ca^2+ab^2)+4(a+b+c)+20].$$It's enough to prove $$2(abc)^2+16\le 4(a+b+c)+15abc+2(ab^2+bc^2+ca^2).\tag{*}$$
After homogenization, it turns out$$4\sqrt{2}(abc)^2+4\sqrt{2}(ab+bc+ca)^3\le 2(a+b+c)\sqrt{(ab+bc+ca)^5}+15abc\sqrt{(ab+bc+ca)^3}+2(ab^2+bc^2+ca^2)\sqrt{(ab+bc+ca)^3},$$or$$\sqrt{(ab+bc+ca)^3}\left[2(a+b+c)(ab+bc+ca)+15abc+2(ab^2+bc^2+ca^2)\right]\ge 4\sqrt{2}(abc)^2+4\sqrt{2}(ab+bc+ca)^3.$$
Let $a=\min{a,b,c}; b=a+u; c=a+v$ where $u,v\ge 0.$
By squaring both side, it is obvious by BW. See Wolfram result.
I'm looking a nice proof for $(*).$
I think RiverLi's approach which verify the yield $ab^2+bc^2+ca^2$ via $p,q,r$ will help but it is much calculate. See his answer for my similar question.
Also, is there another way to solve the starting inequality without full expanding?
I thought of Cauchy-Schwarz inequality $$\sum_{cyc}\frac{a}{a+ab+2}=\sum_{cyc}\frac{a^2(xa+yb+zc)^2}{a(a+ab+2)(xa+yb+zc)^2}\ge \frac{[x(a^2+b^2+c^2)+2(y+z)]^2}{\sum_{cyc}a(a+ab+2)(xa+yb+zc)^2} \ge \frac{2}{3},$$which reduce the degree in some way but I think it is very ugly and it's probably nothing.
Any hint would be desirable.
Best Answer
I hope there might be better solution than this. Thanks to TATA box, it suffices to prove the below.
$$2\sum_{cyc}ab^2 + 15abc+4\sum_{cyc}a \geq 2(abc)^2+16$$
Step 1
By the equality condition, we start with the following.
$$\sum_{cyc} a(2b+c-2)^2 \geq 0$$
$$\sum_{cyc}a \sum_{cyc} ab = 2 \sum_{cyc} a$$
Subtracting these two, and by simplifying, we get the following.
$$ 2\sum_{cyc} a b^2 + 6abc + 4 \sum_{cyc} a \geq 16 $$
Step 2
By step 1, it suffices to prove the following.
$$9abc \geq 2(abc)^2$$
This is trivial since $abc \leq (\frac{ab+bc+ca}{3})^{3/2} <1 < \frac{9}{2}$ by AM-GM. Therefore, the problem is solved.
Note that the equality holds if and only if $(a,b,c)=(2,1,0),(1,0,2),(0,2,1)$.