Can You Prove This Inequality Involving Real Numbers?

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Discussion Overview

The discussion centers around proving the inequality involving positive real numbers: $\dfrac {c}{a+b} +\dfrac {a}{b+c} +\dfrac {b}{c+a}\geq \dfrac {3}{2}$. Participants explore various methods to approach the proof, including calculus and logical reasoning.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose using calculus to find the minimum of the function $f(a,b,c) = \frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{c+a}$ by setting the partial derivatives to zero.
  • Others suggest a non-calculus approach, arguing that the symmetry of the variables $a$, $b$, and $c$ implies that the minimum occurs when $a = b = c$, leading to the conclusion that $f(a,b,c) = \frac{3}{2}$ in that case.
  • One participant notes that this inequality is known as Nesbitt's inequality and provides a reference for various proofs available online.

Areas of Agreement / Disagreement

Participants express differing views on the methods to prove the inequality, with some favoring calculus and others preferring logical reasoning. There is no consensus on a single approach or resolution of the proof.

Contextual Notes

The discussion does not resolve the mathematical steps involved in the proofs, and assumptions regarding the applicability of different methods remain unaddressed.

Albert1
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$ a,b,c \in\mathbb{R}^+ , \,\,\text{Prove:}$

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