Is |x y| < |x| + |y|?

Useful to remember the so-called "Triangle Inequality " for the absolute value:

For any real numbers \(x,y\) \(\,|x+y|\leq|x|+|y|\).
Equality holds if and only if \(xy\geq{0}\), which means, at least one of the two numbers is 0 or they both have the same sign (both positive or both negative).
Or, strict inequality holds if and only if \(xy<0.\)

In our case, we can see that always \(|x-y|=|x+(-y)|\leq|x|+|y|\), because \(|y|=|-y|\).
The strict inequality will hold if and only if \(x(-y)<0\) or \(xy>0.\)

(1) Either \(x\) or \(y\) can be zero, and in addition, we don't know anything about their signs.
Not sufficient.

(2) Obviously sufficient, because we can state with certainty that the given inequality does not hold.

Answer B

There is another property worth remembering:

1. Always true: \(|x+y|\leq{|x|+|y|}\), note that "=" sign holds for \(xy\geq{0}\) (or simply when \(x\) and \(y\) have the same sign);

2. Always true: \(|x-y|\geq{|x|-|y|}\), note that "=" sign holds for \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(|x|>|y|\) (simultaneously).

Thanks for reminding, certainly worth remembering. Just for the practice, we can prove that the second inequality can be deduced from the first one quite easily:

\(|(x-y)+y|\leq|x-y|+|y|\), or \(|x|\leq|x-y|+|y|\), from which we get \(|x-y|\geq{|x|-|y|}.\)

Ans B

Ref to the attached pic

Stmt 1: has two possibilities

Stmt 2 : gives confirmed no

Therefore ans B

Hi.

I am wondering how you got: \(x^2-2xy+y^2<x^2+2|xy|+y^2\)? --> is \(|xy|+xy>0\)?

I understand the concept of squaring both sides of the inequality, but I am wondering why on the left side you are left with -2xy while on the right side you are left with 2|xy|

Thanks!

\(x^2-2xy+y^2<x^2+2|xy|+y^2\) --> cancel x^2+y^2 in both sides: \(-2xy<2|xy|\) --> reduce by 2: \(-xy<|xy|\) --> \(|xy|+xy>0\).

Hope it's clear.

I understand how to reduce on both sides, I'm just wondering why one has absolute value signs (after simplifying) and why one doesn't.

Thanks!

I understand how to reduce on both sides, I'm just wondering why one has absolute value signs (after simplifying) and why one doesn't.

Thanks!

\((|x-y|)^2=(x-y)^2=x^2-2xy+y^2\).
\((|x| + |y|)^2=(|x|)^2+2|x|*|y|+(|y|)^2=x^2+2|xy|+y^2\).

Is |x – y| < |x| + |y|?

(1) y < x

If y < x then |x - y| will always be positive. However, we don't know the signs for x and y so we don't know what |x| and |y| is.
INSUFFICIENT

(2) xy < 0

This means that either x or y is negative and neither x or y are zero.

If we test for positive and negative values of x and y we will find that regardless of what x and y are (as long as one is positive and one is negative) the inequality will always fail (LHS and RHS will always equal one another)
SUFFICIENT

(B)

Bunuel For the last point you mentioned here Always true: \(|x-y|\geq{|x|-|y|}\), note that "=" sign holds for \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(|x|>|y|\) (simultaneously),

What is the 'strict inequality" for this point?

There are so many variations of this with the equalities and Im getting confused. Kindly help.

|x – y| < |x| + |y| is true for all conditions except when
1)x and y are zero or
2)x and y have opposite signs

Statement (1) y < x
Not sufficient

Statement (2) xy < 0
Says x and y have opposite signs and hence the answer is No

(B) is the answer

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