Given the positive real numbers $a,\,b,\,c$ and $x,\,y,\,z$ satisfying the condition: $a+x=b+y=c+z=1$ Prove the inequality $\displaystyle \left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\ge 3$.

Given the positive real numbers $a,\,b,\,c$ and $x,\,y,\,z$ satisfying the condition:

$a+x=b+y=c+z=1$

Prove the inequality $\displaystyle \left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\ge 3$.

My solution:

Rewrite the intended LHS of the inequality strictly in terms of a, b and c, we have:

$x=1-a,\,y=1-b,\,z=1-c$

$\displaystyle \left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)$

$\displaystyle =\left(abc+(1-a)(1-b)(1-c)\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)$

$\displaystyle =\left(abc+1+ab+bc+ca-(a+b+c)-abc\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)$

$\displaystyle =\left(1+ab+bc+ca-(a+b+c)\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)$

$\displaystyle =\left(1-a(1-b)-b(1-c)-c(1-a)\right)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)$

$\displaystyle =\frac{1}{a(1-b)}+\frac{1}{b(1-c)}+\frac{1}{c(1-a)}-1-a(1-b)\left(\frac{1}{b(1-c)}+\frac{1}{c(1-a)}\right)$

$\displaystyle \,\,\,\,\,\,-1-b(1-c)\left(\frac{1}{a(1-b)}+\frac{1}{c(1-a)}\right)-1-c(1-a)\left(\frac{1}{a(1-b)}+\frac{1}{b(1-c)}\right)$

$\displaystyle =\frac{1-b(1-c)-c(1-a)}{a(1-b)}+\frac{1-a(1-b)-c(1-a)}{b(1-c)}+\frac{1-a(1-b)-b(1-c)}{c(1-a)}-3$

$\displaystyle =\frac{1-b+bc-c+ca}{a(1-b)}+\frac{1-a+ab-c+ac}{b(1-c)}+\frac{1-a+ab-b+bc}{c(1-a)}-3$

$\displaystyle =\frac{1-b-c(1-b)+ca}{a(1-b)}+\frac{1-c-a(1-c)+ab}{b(1-c)}+\frac{1-a-b(1-a)+bc}{c(1-a)}-3$

$\displaystyle =\frac{(1-b)(1-c)+ca}{a(1-b)}+\frac{(1-a)(1-c)+ab}{b(1-c)}+\frac{(1-b)(1-a)+bc}{c(1-a)}-3$

$\displaystyle =\frac{1-c}{a}+\frac{c}{1-b}+\frac{1-a}{b}+\frac{a}{1-c}+\frac{1-b}{c}+\frac{b}{1-a}-3$

$\displaystyle \ge 6\sqrt[6]{\frac{1-c}{a}\cdot \frac{c}{1-b}\cdot\frac{1-a}{b}\cdot\frac{a}{1-c}\cdot\frac{1-b}{c}\cdot\frac{b}{1-a}}-3$ (By the AM-GM inequality, with $1-a,\,1-b,\,1-c$ are all positive)

$\displaystyle = 6-3$

$\displaystyle = 3$ (Q.E.D.)