MHB Can You Prove This Inequality Involving Real Numbers?

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Inequality
Click For Summary
The discussion centers on proving the inequality involving positive real numbers a, b, and c: (c/(a+b)) + (a/(b+c)) + (b/(c+a)) ≥ 3/2. Participants explore both calculus and non-calculus methods for proving the inequality, with a focus on the symmetry of the variables. It is suggested that the minimum occurs when a = b = c, leading to a value of 3/2. The problem is identified as Nesbitt's inequality, with references to various proofs available online. The conversation emphasizes the logical approach to the problem without reliance on calculus.
Albert1
Messages
1,221
Reaction score
0
$ 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:
Mathematics news on Phys.org
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)
 

Thread 'Significant figures for inherently bound values'

I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...
View full post »

Similar threads

MHB Proof of Inequality: $|a+b| \leq |a| + |b|$
  • · Replies 8 ·
Replies
8
Views
2K
MHB Can You Solve This Challenging Inequality Problem?
  • · Replies 1 ·
Replies
1
Views
2K
MHB Proving $(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$
  • · Replies 1 ·
Replies
1
Views
1K
MHB Can You Meet the Sine Function Challenge?
  • · Replies 1 ·
Replies
1
Views
1K
MHB Roots of Polynomial: Find $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$
  • · Replies 1 ·
Replies
1
Views
1K
MHB Prove Triangle Inequality: $\frac{a}{\sqrt[3]{4b^3+4c^3}}+...<2$
  • · Replies 1 ·
Replies
1
Views
1K
MHB Find Product: $(a^4+b^4+c^4)$ and $(a^6+b^6+c^6)$
  • · Replies 3 ·
Replies
3
Views
1K
MHB Prove Triangle Inequality: $\sqrt{2}\sin A-2\sin B+\sin C=0$
  • · Replies 2 ·
Replies
2
Views
2K
MHB Prove $\frac{1}{3}\le \frac{u}{v} \le \frac{1}{2}$ with Triangle Sides
  • · Replies 4 ·
Replies
4
Views
1K
MHB Proving $abcd\ge 3$ with $a, b, c, d>0$
  • · Replies 1 ·
Replies
1
Views
1K