# Inequality Problem

• Jan 4th 2011, 01:32 PM
Lamalif
Inequality Problem
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.
• Jan 4th 2011, 03:27 PM
Plato
Suppose that each of $\displaystyle a,~b,~\&~c$ is a positive real numbers.
If $\displaystyle a=b=c$ then $\displaystyle \sqrt{\dfrac{2a}{a+b}}+\sqrt{\dfrac{2b}{b+c}} +\sqrt{\dfrac{2c}{a+c}}=3$
If we have $\displaystyle a<b= c$ then $\displaystyle \sqrt{\dfrac{2a}{a+b}}+\sqrt{\dfrac{2b}{b+c}} +\sqrt{\dfrac{2c}{a+c}}<3$.
WHY?
• Jan 5th 2011, 08:32 AM
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.
• Jan 5th 2011, 08:36 AM
alexmahone
Quote:

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 $\displaystyle 3\sqrt{2}$).
• Jan 5th 2011, 08:51 AM
Lamalif
Quote:

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

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