Can You Prove This Inequality Involving Real Numbers?

[Albert1]

$ a,b,c \in\mathbb{R}^+ , \,\,\text{Prove:}$

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

 

[chisigma]

$ 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 $\chi$ $\sigma$

 

[Albert1]

If without the use of calculus,how can we proceed ?

 

[chisigma]

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

$\chi$ $\sigma$

 

[Albert1]

your analysis is reasonable (Yes)

 

[jakncoke1]

This is also known as Nesbitts inequality.

Here is a link containing a bunch of different proofs for this

Nesbitt's Inequality - AoPSWiki