Prove: One of $a, b$ or $c$ is Sum of Other Two

[anemone]

If $a, b$ and $c$ are three real numbers such that $|a-b|\ge|c|$, $|b-c|\ge|a|$ and $|c-a|\ge|b|$, then prove that one of $a, b$ or $c$ is the sum of the other two.

 

[Mathwebster]

Re: Prove a+b=c

My solution

Spoiler

First off, if two are equal then the third is zero and we're done. So we will assume that none are equal. We will assume that $a \le b \le c$ (without loss of generality). Thus, from the three inequalities given we have

$\begin{eqnarray}
a - b \;& \; \le\; & \;c \;& \;\le b-a\\
b - c\; & \;\le\; &\;a &\; \le c-b\\
a - c \;& \;\le\; &\;b & \;\le c-a
\end{eqnarray}.$

From the right hand side of the first inequality we have $a+c \le b$ and the left hand side of the second inequality we have $b \le a + c$ giving the $b=a+c$

 

[anemone]

Re: Prove a+b=c

If $a, b$ and $c$ are three real numbers such that $|a-b|\ge|c|$, $|b-c|\ge|a|$ and $|c-a|\ge|b|$, then prove that one of $a, b$ or $c$ is the sum of the other two.

Squaring both sides of the inequality of $|a-b|\ge|c|$, we get

$(a-b)^2\ge c^2$

$(a-b)^2-c^2\ge 0$

$(a-b+c)(a-b-c)\ge 0$---(1)

Similarly we have

$(b-c+a)(b-c-a)\ge 0$
$-(b-c+a)(-b+c+a)\ge 0 $
$\rightarrow -(a+b-c)(a-b+c)\ge 0$---(2)

and $(c-a+b)(c-a-b)\ge 0$
$(-c+a-b)(-c+a+b)\ge 0$
$\rightarrow (a-b-c)(a+b-c)\ge 0$---(3)

Multiply these three inequalities yields

$(a-b+c)(a-b-c)(-(a+b-c)(a-b+c))(a-b-c)(a+b-c) \ge 0$

$-(a-b+c)^2(a-b-c)^2(a+b-c)^2 \ge 0$

Apparently, this inequality holds true if either $a-b+c=0 (a+c=b)$ or $a-b-c=0 (b+c=a)$ or $a+b-c=0 (a+b=c)$.