Let $a,b,c,x,y,z\in\mathbb{R}_+$ such that $a+x=b+y=c+z=1.$ Prove the inequality
$$(abc+xyz)\left(\frac1{ay}+\frac1{bz}+\frac1{cx}\right)\geq3$$
I tried using AM-HM to get
$$(abc+xyz)\left(\frac1{ay}+\frac1{bz}+\frac1{cx}\right)\geq9\frac{abc+xyz}{ay+bz+cx}$$
and wrote it as $$\frac{abc+xyz}{ay+bz+cx}=\frac{1-(a+b+c)+(ab+bc+ca)}{(a+b+c)-(ab+bc+ca)}=\frac1{a+b+c-ab-bc-ca}-1$$That is, I have to prove that
$$a+b+c-ab-bc-ca\leq\frac34$$
I tried using Cauchy-Schwarz after this as $$(a(1-b)+b(1-c)+c(1-a))\leq\sqrt{(a^2+b^2+c^2)((1-a)^2+(1-b)^2+(1-c)^2)}$$
but didn't get any idea to simplify it further.
Any help would be appreciated!