Inequality Problem

**Lamalif** · January 4th 2011, 01:32 PM

happy bithday for all.
I hope you that help me to prove this inequalitie:
a,b,c are positiv reel
sqrt(2a/(a+b) + sqrt(2b/(b+c)+ sqrt(2c/(c+a)<3.
Thank you in advance.

**Plato** · January 4th 2011, 03:27 PM

Suppose that each of $a,~b,~\&~c$ is a positive real numbers.
If $a=b=c$ then $\sqrt{\dfrac{2a}{a+b}}+\sqrt{\dfrac{2b}{b+c}} +\sqrt{\dfrac{2c}{a+c}}=3$
If we have $a<b= c$ then $\sqrt{\dfrac{2a}{a+b}}+\sqrt{\dfrac{2b}{b+c}} +\sqrt{\dfrac{2c}{a+c}}<3$ .
WHY?

**Lamalif** · January 5th 2011, 08:32 AM

Thank you Platon.
I have an idea :
a/2<(a+b)/2 and 2/(a+b)<2/a and sqrt(2a/(a+b))<sqrt(2).
by this method we arrive at: sqrt(2a/(a+b))+sqrt(2b/(c+b))+sqrt(2c/(a+c)) <3sqrt(2).
Thank you again.

**alexmahone** · January 5th 2011, 08:36 AM

Originally Posted by Lamalif

Thank you Platon.
I have an idea :
a/2<(a+b)/2 and 2/(a+b)<2/a and sqrt(2a/(a+b))<sqrt(2).
by this method we arrive at: sqrt(2a/(a+b))+sqrt(2b/(c+b))+sqrt(2c/(a+c)) <3sqrt(2).
Thank you again.

Granted, but I thought you wanted to prove that the expression is less than 3 (not $3\sqrt{2}$ ).

**Lamalif** · January 5th 2011, 08:51 AM

Originally Posted by alexmahone

Granted, but I thought you wanted to prove that the expression is less than 3 (not $3\sqrt{2}$ ).

Yes .you are right. But I did not arrive.</span>

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