Question

If a, b ,c are real numbers such that |a-b|>=|c|, |b-c|>=|a|, |c-a|>=|b|, then prove that one of a, b, c is the sum of the other two.

Answer 1

*

Answer 2

*

Answer 3

what does * mean?

Answer 4

just it gives us a reminder when somebody updates this post... :)

Answer 5

put a = b + c
|b-c|>=|a| = |b + c|, this is only true if c<=0, b>=0

Answer 6

we are given a,b,c are real numbers, hence we can plot them on the x axes.
Now due to symmetrical behavior of a,b and c, let can assume a<b<c

here, I am gonna use the fact that |x-y| means distance between x and y.
now,
|a-b| >=|c|
|b-c|>=|a|

adding both,
|a-b| + |b-c| >= |c|+|a|

now since a<b<c,
|a-b| + |b-c| = |c-a|

hence |c-a| >= |c| + |a|,

note that greater is never possible and equal is only possible when a and c have opposite signs !
since a<c, then a must be negative and c must be positive,and b lies somewhere between a and c

now, |a-b| >= |c|
|a-b| + |a| >= |c| + |a|
|a-b| + |a| >= |c-a|
|a| >= |c-a| - |a-b|
|a| >= |c-b|

but we are given |a| <= |c-b|

this means |a| = |c-b|

its easy to force the final conclusion from here.

Answer 7

Thanks @shubhamsrg. I am going to justify why |c-a| - |a-b| = |c-b| in this problem and then finish it from where you left off.

|c-a| = the distance between a and c (call this AC)
|a-b| = the distance between a and b (call this AB)
|c-b| = the distance between b and c (call this BC)

Since we assume a<b<c, AC - AB = BC, which means |c-a| - |a-b| = |c-b| is true.

Skip ahead to |a| = |c-b|, which can be rewritten as |a| = c-b, since b<c means |c-b| = c-b.
|a| = |c - b|
|a| = c - b
|a| + b = c

Since a is negative, |a| = -a.
|a| + b = c
-a + b = c
b = c + a
Or, a+c=b.

Answer 8

nice!

Answer 9

if we multiply all the inequalities 1 , 2 and 3,