# Prove that $(abc+xyz) \left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\geq3$

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!

By AM-GM $$(abc+xyz)\sum_{cyc}\frac{1}{ay}=\sum_{cyc}\left(\frac{zx}{a}+\frac{bc}{y}\right)=\sum_{cyc}\left(\frac{xy}{b}+\frac{ab}{x}\right)=$$ $$=\sum_{cyc}\left(\frac{x(1-b)}{b}+\frac{(1-x)b}{x}\right)=\sum_{cyc}\left(\frac{x}{b}+\frac{b}{x}-x-b\right)=$$ $$=\sum_{cyc}\left(\frac{x}{b}+\frac{b}{x}\right)-3\geq\sum_{cyc}2-3=3.$$