Solving the 1/a + 1/b + 1/c = 3 Equation

  • Context: Undergrad 
  • Thread starter Thread starter c6_viyen_1995
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around solving the equation 1/a + 1/b + 1/c = 3, where a, b, and c are positive real numbers. Participants explore potential solutions and relationships between the variables, as well as implications of a secondary inequality involving roots of the variables.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant identifies a solution where a = b = c = 1, suggesting it may be a known solution.
  • Another participant proposes that if 1/a = 3, then a = 1/3, leading to the conclusion that a, b, c must all be greater than 1/3.
  • There is a suggestion that if a, b, and c are whole numbers, they must all be at least 1, leading to the conclusion that a = b = c = 1 is the only solution in that case.
  • Participants discuss the implications of the second equation involving roots, questioning how it can be "solved" equivalently.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the solutions, with some suggesting specific values while others explore broader conditions. No consensus is reached on a definitive solution or method.

Contextual Notes

The discussion includes assumptions about the positivity of a, b, and c, and the implications of treating them as whole numbers versus real numbers. The relationship between the two equations remains unresolved.

Who May Find This Useful

Readers interested in algebraic equations, inequalities, and mathematical problem-solving may find this discussion relevant.

c6_viyen_1995
Messages
10
Reaction score
0
a,b,c >0

1/a + 1/b + 1/c = 3

How can solve that
⁴√(a³) + ⁴√(b³) + ⁴√(c³) ≥ ³√(a²) + ³√(b²) + ³√(c²)

thank you very much!
 
Mathematics news on Phys.org
Help me!
 
Well, I can see that one solution is a=b=c=1 (you probably already know that, though :wink:).

Past that, I don't know. One number must be greater than one, one less, and one relates to the other two by the second formula.
 
c6_viyen_1995 said:
a,b,c >0

1/a + 1/b + 1/c = 3

How can solve that
⁴√(a³) + ⁴√(b³) + ⁴√(c³) ≥ ³√(a²) + ³√(b²) + ³√(c²)

thank you very much!

If you start with 1/a=3, you find a = 1/3.
With 1/a + 1/b = 3, you can see that a > 1/3, because b > 0.
Due to the symmetry of a and b, that means also that b > 1/3.

With 1/a + 1/b + 1/c = 3, it follows that a,b,c > 1/3.

I suspect that is the intended answer.

Otherwise, if a,b,c are supposed to be whole numbers, it follows that a,b,c ≥ 1.
It follows that then a = b = c = 1, because otherwise we would have fractions.


The second equation can be "solved" equivalently. Do you see how?
 

Similar threads

MHB Proving $(a)^\dfrac{1}{3}+(b)^\dfrac{1}{3}+(c)^\dfrac{1}{3}=0$
  • · 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 Roots of Polynomial: Find $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}$
  • · Replies 1 ·
Replies
1
Views
1K
Graduate !?Puzzle?! -Possible Clue- Split the Difference
  • · Replies 12 ·
Replies
12
Views
2K
Graduate Are these two optimization problems equivalent?
  • · Replies 1 ·
Replies
1
Views
2K
MHB Prove Simultaneous Equations: $a+b+c+d \ne 0$
  • · Replies 2 ·
Replies
2
Views
1K
MHB Solving g(x)-h(x) <0: Find a, b, c, d
  • · Replies 2 ·
Replies
2
Views
1K
MHB Proving $abcd\ge 3$ with $a, b, c, d>0$
  • · Replies 1 ·
Replies
1
Views
1K
Undergrad Solving a System of Nonlinear Equations Symbolically
  • · Replies 9 ·
Replies
9
Views
3K
MHB Challenge Problem #8: 3Σ(1/(√(a^3+1))≥2Σ(√(a+b))
  • · Replies 3 ·
Replies
3
Views
2K