Can You Prove This Inequality Involving Real Numbers?

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SUMMARY

The inequality $\dfrac {c}{a+b} +\dfrac {a}{b+c} +\dfrac {b}{c+a}\geq \dfrac {3}{2}$ for positive real numbers $a, b, c$ is proven using symmetry and logical reasoning. The minimum value occurs when $a = b = c$, yielding a value of $\dfrac{3}{2}$. This approach avoids calculus by leveraging the equality of the variables. The discussion references Nesbitt's inequality, which is a well-known result in inequality theory.

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

$\dfrac {c}{a+b} +\dfrac {a}{b+c} +\dfrac {b}{c+a}\geq \dfrac {3}{2}$
 
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Albert said:
$ 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 $\chi$ $\sigma$
 
If without the use of calculus,how can we proceed ?
 
Albert said:
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

$\chi$ $\sigma$
 
your analysis is reasonable (Yes)
 

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