x<y<0 which of the following inequality is always true?

Question

a.y-x>0 b.x^2 > y^2 c. x-1/x< y-1/y d.|x|<|y|

Answer 1

a), b) and c) are correct. d) incorrect

x and y are both negative

a) is true. x<y<0

x<y subtract y from both sides

x-y <0

multiply by negative one, reversing the inequality sign

-x+y>0 reverse the positions of -x and y: -x+y = y-x

y-x>0

absolute value of x is greater than y. d) is false, it's always false

x^2 is greater than y^2 b) is true.

any number squared is always positive

c) is trickier

x- 1/x < y - 1/y

x < y

divide both sides by xy which is a positive number since x and y are both negative. The product of two negative numbers is always positive

x/xy < y/xy

1/y < 1/x

multiply both sides by -1, which reverses the direction of the inequality sign

-1/y > -1/x

or -1/x < -1/y

add x to the left, y to the right side

x-1/x < y - 1/y is always true c) is correct

check by plugging in some specific numbers for x and y, such as -2<-1<0 and maybe fractions too,

-1/2 < -1/4 < 0. While not proof, it's a good check.