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

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The discussion focuses on proving the inequality involving positive real numbers \(a, b, c\) and \(x, y, z\) under the condition \(a+x=b+y=c+z=1\). The main goal is to establish that \((abc+xyz)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right) \ge 3\). Participants express varying degrees of confidence in the proposed solutions, with some seeking validation from more experienced contributors. The conversation highlights the complexity of the proof and the importance of collaboration in mathematical problem-solving. Overall, the thread emphasizes the need for rigorous verification of mathematical inequalities.
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Given the positive real numbers $a,\,b,\,c$ and $x,\,y,\,z$ satisfy the condition

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

Prove that $$\left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\ge 3$$.
 
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anemone said:
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 $$\left(abc+xyz\right)\left(\frac{1}{ay}+\frac{1}{bz}+\frac{1}{cx}\right)\ge 3--(2)$$.
my solution
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 $(3)=4.5>3$
 
Last edited:
Albert said:
my solution
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:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$= 6-3$$

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

Equality occurs when $$\frac{1-c}{a}=\frac{c}{1-b}=\frac{1-a}{b}=\frac{a}{1-c}=\frac{1-b}{c}=\frac{b}{1-a}.$$
 
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