# Show that -1 < or equal 2(x-y)(1+xy)/(1+x^2)(1+y^2) < or equal 1?

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### 1 Answer

You need to prove the following inequality:

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

Since `1+x^2>0` and `1+y^2>0` , hence `((1+x^2)(1+y^2))> 0` and you may multiply both sides by `((1+x^2)(1+y^2))` keeping the direction of the inequality, such that:

`-((1+x^2)(1+y^2)) =< 2x + 2x^2y - 2y - 2xy^2 <= ((1+x^2)(1+y^2))`

`-(1 + y^2 + x^2 + x^2*y^2) =< 2x + 2x^2y - 2y - 2xy^2 <= (1 + y^2 + x^2 + x^2*y^2)`

`-(1 + y^2 + x^2 + x^2*y^2) - 2x + 2y =<2x^2y - 2xy^2 <= (1 + y^2 + x^2 + x^2*y^2) `

`-1 - 2x - x^2 - y^2 - 2y - 1 + 1 - x^2*y^2 =< 2x^2y - 2xy^2 <= (1 + y^2 + x^2 + x^2*y^2) `

`-((x+1)^2 + (y+1)^2- 1 + x^2*y^2) =< 2x^2y - 2xy^2 <= (1 + y^2 + x^2 + x^2*y^2) `

`-((x+1)^2 + (y+1)^2 - 1 + x^2*y^2) =< 0 <= 1 + y^2 + x^2 + x^2*y^2 + 2xy^2 - 2x^2y `

`-((x+1)^2 + (y+1)^2 + x^2*y^2) =< 0 <= y^2 + x^2 + x^2*y^2 + 2xy^2 - 2x^2y `

`-((x+1)^2 + (y+1)^2 + x^2*y^2) =< 0 <= y^2 + x^2 + 2x^2*y^2 + 2xy^2 - 2x^2y + 2xy-2xy`

`-((x+1)^2 + (y+1)^2 + x^2*y^2) =< 0 <= (x+y)^2 + 2x^2*y^2 + 2xy(y - x - y)`

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

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

If y is negative, hence `y - 1` is negative `=> 2x^2*y(y - 1` ) rests positive.

If y is positive and `y>1` , hence `y-1>0 => 2x^2*y(y - 1)` rests positive

Hence, the inequality `0 <= (x+y)^2 + 2x^2*y(y - 1)` is valid.

Since `(x+1)^2 + (y+1)^2 + x^2*y^2 >= 0` , hence `-((x+1)^2 + (y+1)^2 + x^2*y^2) =< 0`

**Hence, checking if the inequality is valid yields that `-1 =< (2(x-y)(1+xy))/((1+x^2)(1+y^2)) <= 1` holds.**