Given three non-negative real numbers $a$, $b$ and $c$. The sum of the modules of their pairwise differences is equal to $1$, i.e. $|a- b| + |b -c| + |c -a| = 1$. What can the sum $a + b + c$ be equal to?
Problem
Source: 2023 South Russian Girls MO - Assara Seniors p5
Tags: algebra
sqing
02.03.2024 06:44
Let $ a,b,c\geq 0 $ and $|a- b| + |b -c| + |c -a| = 1$. Prove that $$a + b + c\geq \frac{1}{2}$$
parmenides51 wrote:
Given three non-negative real numbers $a$, $b$ and $c$. The sum of the modules of their pairwise differences is equal to $1$, i.e. $|a- b| + |b -c| + |c -a| = 1$. What can the sum $a + b + c$ be equal to?
SBLNuclear17
02.03.2024 07:05
sqing wrote: Let $ a,b,c\geq 0 $ and $|a- b| + |b -c| + |c -a| = 1$. Prove that $$a + b + c\geq \frac{1}{2}$$ To solve this problem, lets first assume that $x\leq y\leq z$, and that $x=0$. Here is a graph of Y and Z, when x is 0. From the picture, We can clearly see that the minimum sum of y and z is $\frac12$. Problem Solved.
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Tem8
02.03.2024 07:20
WLOG $a \ge b \ge c$. Let $x=a-b$ and $y=b-c$. Then $|a- b| + |b -c| + |c -a| = 2(x+y)=1$, and $x+y=0.5$.
$a+b+c = 3c + 2x + y = 3c + y + 0.5 \ge 0.5$