# compare P and Q

#### Albert

##### Well-known member
$a,b,c,d,m,n >0$

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

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

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

#### caffeinemachine

##### Well-known member
MHB Math Scholar
$a,b,c,d,m,n >0$

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

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

$compare\,\, P \,\,and \,\,Q$
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 $Q\geq P$.

#### anemone

##### MHB POTW Director
Staff member
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 $\displaystyle \frac{a}{b}=\frac{c}{d}$.

Last edited:

#### Albert

##### Well-known member
$a,b,c,d,m,n >0$

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

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

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