Understand the problem

Determine the minimal and maximal values the expression $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|}$ can take, where $a,b,c$ are real numbers.
Source of the problem
Israel MO 2018, Problem 3
Topic
Algebra, Inequality
Difficulty Level
6/10
Suggested Book
Excursion in Mathematics by Bhaskarcharya Prathisthan

Start with hints

Do you really need a hint? Try it first!

Given the expressions, what inequality comes to your mind first? The triangle inequality right? |x| + |y| \( \geq \) |x+y|. Can you use this inequality to get a maximum bound?
Maximum Bound: Observe that the maximum bound is got by the triangle inequality as explained. $|a+b| \le |a|+|b|$
$|b+c| \le |b|+|c|$  
$|a+c| \le |a|+|c|$ We get, $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|} \le\frac{|a|+|b|+|b|+|c|+|a|+|c|}{|a|+|b|+|c|}=2$ Never forget to mention the equality case: a = b = c is the equality case.  
What about the minimum inequality? The idea is that you can observe that keeping the denominator constant, can you reduce the numerator. Let’s take the case of the |a| = |b| = |c| = 1. Now, the expression is maximized when a = b = c = 1 or -1. So, obviously one must be positive or two must be negative or vice-versa. In either case, we get \( \frac{2}{3} \). Okay, then maybe we need to deal with the signs and stuff to get a hold on the minimum. Let’s fix the signs of a,b,c then, we can break the bonds of the modulus. Let’s proceed to the next hint.

Let $a$, $b$, and $c$ be arbitrary real numbers, not all of them equal $0$. By flipping signs, we can assume that at least two of $a$, $b$, and $c$ are non-negative. Actually, without loss of generality, we can assume that $a, b\geq 0$. \( 3|a + b| + 3|b + c| + 3|c + a| \geq 3a + 3b + 3|b+c| + 3|a+c| \geq 2(a+b) + (a + |a + c|) + (b + |b + c|) \) \( = 2(|a| + |b|) + (|-a| + |a + c|) + (|-b| + |b + c|) \geq 2(|a| + |b|) + |c| + |c| = 2|a| + 2|b| + 2|c| \) We have proved that the minimum possible value of $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|}$ is $\frac{2}{3}$. The minimum is $\frac{2}{3}$, which is attained for $a = b = 1$$c = -1$.  

Watch video

Connected Program at Cheenta

Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

Similar Problems

Functional Equation Problem from SMO, 2018 – Question 35

Try this problem from Singapore Mathematics Olympiad, SMO, 2018 based on Functional Equation. You may use sequential hints if required.

Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and the greatest integer.

Arithmetic sequence | AMC 10A, 2015 | Problem 7

Try this beautiful problem from Algebra: Arithmetic sequence from AMC 10A, 2015, Problem. You may use sequential hints to solve the problem.

Series and sum | AIME I, 1999 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.

Inscribed circle and perimeter | AIME I, 1999 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Rectangles and sides.

Problem based on Cylinder | AMC 10A, 2015 | Question 9

Try this beautiful problem from Mensuration: Problem based on Cylinder from AMC 10A, 2015. You may use sequential hints to solve the problem.

Cubic Equation | AMC-10A, 2010 | Problem 21

Try this beautiful problem from Algebra, based on the Cubic Equation problem from AMC-10A, 2010. You may use sequential hints to solve the problem.

Median of numbers | AMC-10A, 2020 | Problem 11

Try this beautiful problem from Geometry based on Median of numbers from AMC 10A, 2020. You may use sequential hints to solve the problem.

LCM and Integers | AIME I, 1998 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1998, Problem 1, based on LCM and Integers.

Problem on Fraction | AMC 10A, 2015 | Question 15

Try this beautiful Problem on Fraction from Algebra from AMC 10A, 2015. You may use sequential hints to solve the problem.