Question

Prove that for every x>0,x/1+x^2<tan^-1<x?

Answer 1

For `x > 0` we have that `1+x^2 > 1` so that:

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

Consider now the expression:

`(2x^2)/(1+x^2)^2`

as for `x != 0` this quantity is always positive we also have:

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

and simplifying:

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

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

Putting the two inequalities together:

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

Integrating over the same interval preserves the inequality, so:

`int_0^x(1-t^2)/(1+t^2)^2 dt < int_0^x (dt)/(1+t^2) < int_0^x dt`

And as:

`int_0^x dt = x`

`int_0^x (dt)/(1+t^2) = arctanx`

`int_0^x(1-t^2)/(1+t^2)^2 dt = x/(1+x^2)`

the inequality is proven.