Compare P and Q: $a, b, c, d, m, n > 0$

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Discussion Overview

The discussion revolves around comparing two expressions, \( P \) and \( Q \), defined in terms of positive variables \( a, b, c, d, m, n \). The focus is on demonstrating the relationship between these two expressions using mathematical inequalities, specifically the AM-GM inequality.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Participants define \( P \) as \( \sqrt{ab} + \sqrt{cd} \) and \( Q \) as \( \sqrt{ma+nc} \times \sqrt{\frac{b}{m}+\frac{d}{n}} \).
  • One participant proposes that to compare \( P \) and \( Q \), it suffices to show that \( Q^4 \geq P^4 \), which can be transformed into an inequality involving \( (m/n)^2(ad)^2 + (n/m)^2(bc)^2 \geq 2abcd \).
  • This participant claims that the inequality holds true by the AM-GM inequality, concluding that \( Q \geq P \).
  • Another participant presents a different approach, showing that \( Q^2 \) can be expressed in a way that leads to \( Q^2 \geq P^2 \) through a similar application of the AM-GM inequality, also concluding that \( Q \geq P \).

Areas of Agreement / Disagreement

Participants appear to agree on the conclusion that \( Q \geq P \) based on their respective applications of the AM-GM inequality, though the specific methods and transformations used differ.

Contextual Notes

The discussion relies on the assumptions that all variables \( a, b, c, d, m, n \) are positive, and the validity of the AM-GM inequality is taken as a given without further exploration of conditions or limitations.

Albert1
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$a,b,c,d,m,n >0$

$ P=\sqrt {ab}+\sqrt {cd}$

$Q=\sqrt {ma+nc} \times \sqrt {\dfrac {b}{m}+\dfrac {d}{n}}$

$compare\,\, P \,\,and \,\,Q$
 
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Albert said:
$a,b,c,d,m,n >0$

$ P=\sqrt {ab}+\sqrt {cd}$

$Q=\sqrt {ma+nc} \times \sqrt {\dfrac {b}{m}+\dfrac {d}{n}}$

$compare\,\, P \,\,and \,\,Q$
We show that $Q^4\geq P^4$.

This is same as showing that:

$(m/n)^2(ad)^2+(n/m)^2(bc)^2\geq 2abcd$.

This is clearly true by the AM-GM inequality.

Therefore $Q\geq P$.
 
My solution:
By applying Cauchy-Schwarz Inequality theorem to Q, we have:

$$Q=\sqrt {ma+nc} \times \sqrt {\dfrac {b}{m}+\dfrac {d}{n}} \ge \sqrt{ma}\sqrt{\frac{b}{m}}+\sqrt{nc}\sqrt{\frac{d}{n}} \ge \sqrt{ab}+\sqrt{cd}= P$$

and equality holds iff $\displaystyle \frac{a}{b}=\frac{c}{d}$.
 
Last edited:
Albert said:
$a,b,c,d,m,n >0$

$ P=\sqrt {ab}+\sqrt {cd}$

$Q=\sqrt {ma+nc} \times \sqrt {\dfrac {b}{m}+\dfrac {d}{n}}$

$compare\,\, P \,\,and \,\,Q$

$Q^2=ab+cd+\dfrac {mad}{n}+\dfrac {nbc}{m}\geq ab+cd+2\sqrt {\dfrac {mad\times nbc}{mn}}=ab+cd+2\sqrt {abcd}=P^2$
$\therefore Q\geq P$
 

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