Inequality challenge for all positive (but not zero) real a, b and c

Click For Summary
SUMMARY

The discussion centers on proving the inequality $$\frac{ab}{a+b+ab}+\frac{bc}{b+c+bc}+\frac{ca}{c+a+ca}\le \frac{a^2+b^2+c^2+6}{9}$$ for all positive real numbers \(a\), \(b\), and \(c\) where \(a, b, c \neq 0\). Participants confirm the validity of the approach taken by user Albert, indicating a consensus on the method used to tackle the problem. The inequality is established as a significant challenge in mathematical analysis, particularly in the realm of inequalities involving multiple variables.

PREREQUISITES
  • Understanding of algebraic inequalities
  • Familiarity with positive real numbers
  • Knowledge of mathematical proofs
  • Experience with symmetric functions
NEXT STEPS
  • Study techniques for proving inequalities in multiple variables
  • Explore the Cauchy-Schwarz inequality and its applications
  • Investigate the use of AM-GM inequality in algebraic proofs
  • Learn about symmetric sums and their properties in inequalities
USEFUL FOR

Mathematicians, students of advanced algebra, and anyone interested in the field of inequalities and mathematical proofs will benefit from this discussion.

anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Prove $$\frac{ab}{a+b+ab}+\frac{bc}{b+c+bc}+\frac{ca}{c+a+ca}\le \frac{a^2+b^2+c^2+6}{9}$$ for all positive real $a,\,b$ and $c$ and $a,\,b,\,c\ne 0$.
 
Mathematics news on Phys.org
anemone said:
Prove $$\frac{ab}{a+b+ab}+\frac{bc}{b+c+bc}+\frac{ca}{c+a+ca}\le \frac{a^2+b^2+c^2+6}{9}$$ for all positive real $a,\,b$ and $c$ and $a,\,b,\,c\ne 0$.

Hint:

Note that $$\frac{ab}{a+b+ab}=\frac{1}{\frac{1}{a}+\frac{1}{b}+1}$$.
 
my solution:
using $AP\geq HP$
We have :$\\
\dfrac{ab}{a+b+ab}\leq \dfrac{a+b+1}{9}---(1)\\
\dfrac{bc}{b+c+bc}\leq \dfrac{b+c+1}{9}---(2)\\
\dfrac{ca}{c+a+ca}\leq \dfrac{c+a+1}{9}---(3)\\
(1)+(2)+(3):\dfrac {ab}{a+b+ab}+\dfrac {bc}{b+c+bc}+\dfrac {ca}{c+a+ca}\leq\dfrac{2a+2b+2c+3}{9}\leq \dfrac{a^2+1+b^2+1+c^2+1+3}{9}= \dfrac{a^2+b^2+c^2+6}{9}$
(using $AP\geq GP$)
 
Well done Albert!(Cool) That is how I approached the problem as well!
 

Similar threads

MHB Solving for $(a,\,b,\,c)$ with Powers of 2
  • · Replies 1 ·
Replies
1
Views
1K
MHB Prove Geometric Sequence with $(a,b,c)$
  • · Replies 1 ·
Replies
1
Views
1K
MHB Prove that the sum of 6 positive integers is a composite number
  • · Replies 1 ·
Replies
1
Views
1K
MHB Inequality Challenge: Prove $\ge 0$ for All $a,b,c$
  • · Replies 2 ·
Replies
2
Views
2K
MHB Prove Inequality: $\sqrt{ab}+\sqrt{cd}\le \sqrt{(a+d)(b+c)}$
  • · Replies 1 ·
Replies
1
Views
1K
MHB Positive Integers: Evaluate $a+b+c$
  • · Replies 1 ·
Replies
1
Views
1K
MHB Inequality with positive real numbers a and b
  • · Replies 1 ·
Replies
1
Views
1K
MHB Prove Inequality for $a,\,b,\,c$: $9abc\ge7(ab+bc+ca)-2$
  • · Replies 3 ·
Replies
3
Views
2K
MHB Can You Solve This Challenging Inequality Problem?
  • · Replies 1 ·
Replies
1
Views
2K
MHB Prove the inequality (a^3-c^3)/3≥abc((a-b)/c+(b-c)/a)
  • · Replies 3 ·
Replies
3
Views
2K