# Given x >0 , show inequality ln(1+x^2)< x arctg x < x^2 ?

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### 1 Answer

You need to move all terms to the left side of inequality, such that:

`ln(1 + x^2) - x*arctan x - x^2 < 0`

You need to prove that the function `f(x) = ln(1 + x^2) - x*arctan x - x^2` is negative, if` x>0` , hence, you should test if its derivative is negative, such that:

`f(x) = ln(1 + x^2) - x*arctan x - x^2`

`f'(x) = (2x)/(1 + x^2) - arctan x - x/(1 + x^2) - 2x`

`f'(x) = x/(1 + x^2) - arctan x - 2x`

You cannot tell if derivative is negative if x>0, hence, you should continue the process and you need to differentiate again, such that:

`f''(x) = (x'(1 + x^2) - x*(1 + x^2)')/((1 + x^2)^2) - 1/(1 + x^2) - 2`

`f''(x) = (1 + x^2 - 2x^2)/((1 + x^2)^2) - 1/(1 + x^2) - 2`

`f''(x) = (1 - x^2 - 1 - x^2 - 2(1 + x^2)^2)/((1 + x^2)^2) `

Reducing duplicate terms yields:

`f''(x) = -2(x^2 + (1 + x^2)^2)/((1 + x^2)^2)< 0 => {(-2<0),(x^2 + (1 + x^2)^2 > 0),((1 + x^2)^2 > 0):}`

Since `f''(x) < 0` for `x > 0` , hence f'(x) < 0, thus `f(x)<0.`

Replacing `ln(1 + x^2) - x*arctan` `x - x^2` for `f(x)` yields:

`ln(1 + x^2) - x*arctan x - x^2 < 0`

**Hence, testing if `ln(1 + x^2) - x*arctan x - x^2 < 0 ` holds, the answer is affirmative**.