# Inequality .Prove that y =< 1/2 if y = x/(x^2+1) .

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### 2 Answers

The problem provides the following information, such that:

`{(y<= 1/2),(y = x/(x^2+1)):} => x/(x^2+1) <= 1/2`

`x/(x^2+1) -1/2 <= 0 => (2x - (x^2 + 1))/(x^2 + 1) <= 0`

`(2x - x^2 - 1)/(x^2 + 1) <= 0 `

Factoring out -1 to numerator yields:

`-(x^2 - 2x + 1)/(x^2 + 1) <= 0 => -(x - 1)^2/(x^2 + 1) <= 0`

Since `(x - 1)^2 >= 0` and `(x^2 + 1) > 0` , hence `(x - 1)^2/(x^2 + 1) >= 0` , thus `-(x - 1)^2/(x^2 + 1) <= 0.`

**Hence, testing if `y <= 1/2` , under the given conditions, yields **`-(x - 1)^2/(x^2 + 1) <= 0.`

For solving the inequality, we'll substitute y by the expression in x:

x/(x^2+1) =< 1/2

We'll subtract 1/2 both sides:

x/(x^2+1) - 1/2 =< 0

To compute the difference between 2 ratios, we'll compute the LCD.

LCD = 2(x^2+1)

We'll multiply the first fraction by 2 and the second fraction by (x^2+1).

The inequality will become:

2x - (x^2+1) =< 0

We'll remove the brackets and we'll re-arrange the terms:

-x^2 + 2x - 1 =< 0

We'll multiply by -1:

x^2 - 2x + 1 > = 0

The expression is the result of expanding the square (x-1)^2:

(x-1)^2 > = 0

The square is always positive for any value of x.

The expresion is zero for x - 1 = 0

x = 1.