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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

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

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

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

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

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

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

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

- Mar 10, 2012

- 834

We show that $Q^4\geq P^4$.

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

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

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

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 $Q\geq P$.

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- #3

- Feb 14, 2012

- 3,755

My solution:

By applying Cauchy-Schwarz Inequality theorem to Q, we have:

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

and equality holds iff $\displaystyle \frac{a}{b}=\frac{c}{d}$.

By applying Cauchy-Schwarz Inequality theorem to Q, we have:

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

and equality holds iff $\displaystyle \frac{a}{b}=\frac{c}{d}$.

Last edited:

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- Jan 25, 2013

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

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

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

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

$\therefore Q\geq P$