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inequality

Albert

Well-known member
Jan 25, 2013
1,225
$ a,b,c \in\mathbb{R}^+ , \,\,\text{Prove:}$

$\dfrac {c}{a+b} +\dfrac {a}{b+c} +\dfrac {b}{c+a}\geq \dfrac {3}{2}$
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
$ a,b,c \in\mathbb{R}^+ , \,\,\text{Prove:}$

$\dfrac {c}{a+b} +\dfrac {a}{b+c} +\dfrac {b}{c+a}\geq \dfrac {3}{2}$
Setting $\displaystyle f(a,b,c) = \frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{c+a}$ find the points of minimum imposing $\displaystyle \frac{\partial f}{\partial a} = \frac{\partial f}{\partial b} = \frac{\partial f}{\partial c} = 0$ and then compute f(*,*,*) in these points...


Kind regards


$\chi$ $\sigma$
 

Albert

Well-known member
Jan 25, 2013
1,225
If without the use of calculus,how can we proceed ?
 

chisigma

Well-known member
Feb 13, 2012
1,704
If without the use of calculus,how can we proceed ?
As 'pure logical' approach to the problem You can consider that the role of the variable a, b and c is exactly the same, i.e. You can swap them and nothing changes. That means that the minimum of f(*,*,*) is when a=b=c and in that case is f(*,*,*)= 3/2...

Kind regards

$\chi$ $\sigma$
 

Albert

Well-known member
Jan 25, 2013
1,225
your analysis is reasonable (Yes)
 

jakncoke

Active member
Jan 11, 2013
68