To prove y < 1/2 if y = x/(x^2+1).

x/(x^2+1) = y

y' = { (x)'(x^2+1) - x(x^2+1)'}/(x^2+1)

y' = 0 gives: {x^2+1 - x(2x)}/(x^2+1)^2. = 0

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

Therefore x^2 = 1, or x = 1 or x= -1.

y'' = {(1-x^2)' ((x^2+1) - ((1-x^2){(1+x^2)^2}/(1+x^2)^4.

y" = {-2x(x^2+1) - 2(1-x^2)(1+x^2)}/(1+x^2)

{y'' at x= 1 } = {-2(1+1) - 0*(..)}/(1+1)^4 < 0.

Therefore y (x) maximumm when x=1 Or y(1) = 1/(1^2+1) = 1/2.

So y(x) < y(1) = 1/2.

Or y < 1/2 . if y = x/(x^2+1).

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.