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.
Problem
Source: Israel National Olympiad 2018 Q3
Tags: absolute value, minimum value, maximum value, maximum and minimum, algebra, Inequality
bel.jad5
08.08.2019 01:25
Cuubic wrote: 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. Obviously: $max=2$.
Mixer_V
08.08.2019 01:31
Since $|a+b| \le |a|+|b|$,
$|b+c| \le |b|+|c|$ and
$|a+c| \le |a|+|c|$, we have
$\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|} \le\frac{|a|+|b|+|b|+|c|+|a|+|c|}{|a|+|b|+|c|}=2$
BlazingMuddy
08.08.2019 02:44
The minimum is $\frac{2}{3}$, which is attained for $a = b = 1$, $c = -1$.
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 $a, b\geq 0$. Then,
\begin{align*}
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|
\end{align*}
We have proved that the minimum possible value of $\frac{|a+b|+|b+c|+|c+a|}{|a|+|b|+|c|}$ is $\frac{2}{3}$.
sqing
13.08.2019 06:02
Cuubic wrote: 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. https://artofproblemsolving.com/community/c6h1568635p9643025