MHB Can You Prove This Inequality Involving Real Numbers?

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Inequality
AI Thread 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 'Trigonometry problem of interest'

Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
View full post »

Similar threads

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

Hot Threads

Recent Insights

Back
Top