Let $a \ge b\ge c\ge 0$ such that $ab+bc+ca=3.$ Find minimum $$T= a+b+(a-b)(c-b).$$ For $a=b=c=1,$ T get minimal value is equal to $2.$ Hence, I tried to prove $a+b+(a-b)(c-b)\ge 2.$
My ugly proof using SOS.
After homogenize, we need to prove $$(a+b)\sqrt{3(ab+bc+ca)} \ge 5b(a+c)-ac-3b^2,$$ or $$3(a+b)^2(ab+bc+ca) \ge [5b(a+c)-ac-3b^2]^2.$$ The last inequality is equivalent to $$\left (b(a-b)+\frac{24ba+28bc+4ac+7c^2}{12}\right)(a-3b+2c)^2+\frac{c(32a^2+14c^2+48ab+21bc+29ca)(a+b-2c)}{12}\ge 0.$$
Hope to see simpler ideas. Thanks.
Proof.
We will prove$$a+b+(a-b)(c-b)\ge 2. \tag{*}$$
By given hypothesis, $3\ge 3c^2 \implies c\in [0;1].$
The OP can be rewritten as $$a+b+ac-ab+b^2-bc-2\ge 0\iff a(1+2c)+b^2+b-5\ge 0$$ By replacing $a=\dfrac{3-bc}{b+c},$ it is enough to prove$$-2bc^2+c(b^2+1)+b^3+b^2-5b+3\ge 0,$$which is equivalent to $$c[(b-c)^2+1-c^2]+(b-1)^2(b+3)\ge 0.$$ Hence $(*)$ is proven.
The minimal value is equal to $2$ when $a=b=c=1$ or $a=3,b=1,c=0.$