If $a-b > \sqrt{(b+c-a)^2}$ which of the following must be true?

Recognize that the right side is positive, so A must be greater than B.

Unfortunately, that is not one of the answer choices. Too bad.

For a given result on the right side, there are two ways to achieve this.

If B+C>A, then right side equals B+C-A.

For example, B=2, C=4 and A=5. So right side would equal 1.

So, A-B >B+C-A. Unfortunately, this doesn't help us since any manipulation of this won't yield any of the answer choices. Too bad again.

However, right side can also = 1 if B+C-A= -1, since squaring -1=1 and square root of 1=1.
B could=2, C=4 and A=7 for example.
But squaring this and taking the square root loses the relationship between A,B and C.

Fortunately B+C-A can be rewritten as A-B-C to yield +1 in this example.

So, in this situation, A-B>A-B-C

Subtract A and add B to both sides yields 0>-C. This is only true if
C>0 D

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

Re: If $a-b > \sqrt{(b+c-a)^2}$ which of the following must be true?

by regor60 » Wed Nov 17, 2021 1:00 pm

Another way, probably better.

Rearrange B+C-A to be C-(A-B)

Call (A-B)= X

Now X > $$\sqrt{\left(C-X\right)}^2$$

Square both sides

X $$^2$$ > $$^{\left(C-X\right)^2}$$

So $$^{\left(X^2\right)}$$ > $$^{\left(C^2\right)}$$ -2XC+ $$^{\left(X^2\right)}$$

Subtract $$^{\left(X^2\right)}$$ from both sides

0>C(C-2X)

This can only be true if either
C>0 and (C-2X)<0
Or
C<0 and (C-2X)>0

Let's examine the second choice first

For C-2X >0 C must be> 2X

2X=2(A-B) from our substitution.

But A is greater than B because the right side of the problem statement is positive always, so 2(A-B) is positive.

But for C to be greater than 2(A-B) means therefore that C>0.

But C can't be >0 per the condition, creating a contradiction, so the second choice is eliminated, leaving only the first choice C>0 D