# Understand the problem

Determine the minimal and maximal values the expression can take, where 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.

We get, Never forget to mention the equality case: a = b = c is the equality case.

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

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