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

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SUMMARY

The discussion focuses on comparing two mathematical expressions, P and Q, defined as \( P = \sqrt{ab} + \sqrt{cd} \) and \( Q = \sqrt{ma + nc} \times \sqrt{\frac{b}{m} + \frac{d}{n}} \). It is established that \( Q^4 \geq P^4 \) can be derived from the AM-GM inequality, leading to the conclusion that \( Q \geq P \) for all positive values of \( a, b, c, d, m, n \). The proof involves manipulating the expressions to demonstrate that \( Q^2 \) is greater than or equal to \( P^2 \).

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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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