Prove that if x and y are distinct and positive real numbers then (x/y) + (y/x) > 2. discuss the conditions (on x and y) under which this inequality becomes an equality.

Respuesta :

1.

[tex] \dfrac{x}{y}+\dfrac{y}{x} > 2|\cdot xy\\
x^2+y^2>2xy\\
x^2-2xy+y^2>0\\
(x-y)^2>0 [/tex]

[tex] (x-y)^2>0 [/tex] is true for any two real numbers such that [tex] x\not =y [/tex], all the more for positive ones.

2.

[tex] x=y [/tex]

The (x/y) + (y/x) > 2 is always correct for real numbers except x = y then inequality becomes an equality.

What is a number system?

The number system is a way to represent or express numbers.

All number that we can think of is a real number except complex numbers.

The Number System includes any of the several sets of symbols and the rules for using them to represent numbers.

A decimal number is a very common number that we use frequently.

Given two real numbers x and y

(x/y) + (y/x) > 2

Taking LCM

(x² + y²)/xy > 2

x² + y² > 2xy

x² + y² - 2xy > 0

(x - y)² > 0

This is always correct for two real numbers except x = y at this condition x - y = 0  and 0 = 0 so inequality converts into equality.

Hence, the (x/y) + (y/x) > 2 is always correct for real numbers except x = y then inequality becomes an equality.

For more about the number system,

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