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Please prove the following inequality

Albert

Well-known member
Jan 25, 2013
1,225
a,b,c,d > 0 , please prove :

$ \sqrt{a+b+c+d} \geq \dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d}}{2}$
 
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jakncoke

Active member
Jan 11, 2013
68
Now

$\displaystyle \left(\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}}{2} \right)^2 = \frac{a+b+c+d+2\sqrt{ab}+2\sqrt{ac} + 2\sqrt{ad} + 2\sqrt{bc} + 2\sqrt{cd} + 2\sqrt{bd}}{4}$

use the AM-GM inequality which says $\displaystyle \frac{a+b}{2} \geq \sqrt{ab}$ or $a + b\geq 2 \sqrt{ab}$

so we get

$a+b \geq 2 \sqrt{ab}$ , $a + c \geq 2 \sqrt{ac}$ , etc... for all the square root terms, we get

$4a+4b+4c+4d \geq 2\sqrt{ab}+2\sqrt{ac} + 2\sqrt{ad} + 2\sqrt{bc} + 2\sqrt{cd} + 2\sqrt{bd} + a + b + c + d$

and dividing left and right by 4

$\displaystyle a+b+c+d \geq \frac{a+b+c+d+2\sqrt{ab}+2\sqrt{ac} + 2\sqrt{ad} + 2\sqrt{bc} + 2\sqrt{cd} + 2\sqrt{bd}}{4}$ and since the square root of the right hand side is $\displaystyle \left(\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}}{2} \right)$

we get

$\displaystyle \sqrt{a+b+c+d} \geq \frac{\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}}{2}$
 
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Albert

Well-known member
Jan 25, 2013
1,225
By Cauchy-Schwarz inequality:

$({1}^{2}+{1}^{2}+{1}^{2}+{1}^{2})({\sqrt{a}}^{2}+{\sqrt{b}}^{2}+{\sqrt{c}}^{2}+{\sqrt{b}}^{2})

\geq (\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2$


$4(a+b+c+d)\geq (\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2 $

$ \therefore \sqrt{a+b+c+d}\geq \dfrac{(\sqrt{a}+\sqrt{b}++\sqrt{c}++\sqrt{d})}{2}$
 

jakncoke

Active member
Jan 11, 2013
68
By Cauchy-Schwarz inequality:

$({1}^{2}+{1}^{2}+{1}^{2}+{1}^{2})({\sqrt{a}}^{2}+{\sqrt{b}}^{2}+{\sqrt{c}}^{2}+{\sqrt{b}}^{2})

\geq (\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2$


$4(a+b+c+d)\geq (\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d})^2 $

$ \therefore \sqrt{a+b+c+d}\geq \dfrac{(\sqrt{a}+\sqrt{b}++\sqrt{c}++\sqrt{d})}{2}$
this is a very elagant and simple proof.