Proving the Relation between $a,\,b,\,c$ and $x,\,y,\z$

[anemone]

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

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

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

 

[Albert1]

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

$a+x=b+y=c+z=1---(1)$

Prove that \(\displaystyle \left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\ge 3--(2)\).

my solution

Spoiler

expansion of $(2)$ we have :
$\dfrac{xz}{a}+\dfrac{xy}{b}+\dfrac{yz}{c}+\dfrac{ab}{x}+\dfrac{bc}{y}+\dfrac{ac}{z}\\
\geq 6\sqrt[6]{abcxyz}---(3)$
using $AP\geq GP$ on $ (1)(3)$
equality will happen when $a=b=c=x=y=z=0.5$
and $(3)\geq 6\times 0.5=3$
you may check another point with $a=b=c=\dfrac {1}{3} ,x=y=z=\dfrac {2}{3}$
and get $(3)=4.5>3$

 

[anemone]

my solution

Spoiler

expansion of $(2)$ we have :
$\dfrac{xz}{a}+\dfrac{xy}{b}+\dfrac{yz}{c}+\dfrac{ab}{x}+\dfrac{bc}{y}+\dfrac{ac}{z}\\
\geq 6\sqrt[6]{abcxyz}---(3)$
using $AP\geq GP$ on $ (1)(3)$
equality will happen when $a=b=c=x=y=z=0.5$
and $(3)\geq 6\times 0.5=3$
you may check another point with $a=b=c=\dfrac {1}{3} ,x=y=z=\dfrac {2}{3}$
and get $(3)=4.5>3$

Thanks for participating, Albert...I guess your solution works, but I am not that certain, I would be happy and grateful if someone who is more expert can give their two cents here...(Mmm)

My solution:

Spoiler

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.)

Equality occurs when \(\displaystyle \frac{1-c}{a}=\frac{c}{1-b}=\frac{1-a}{b}=\frac{a}{1-c}=\frac{1-b}{c}=\frac{b}{1-a}.\)