Doubts about proving an inequality with arctangent

[dunkelheit]

Prove that [MATH]\arctan x > \frac{x}{1+x^2}[/MATH] for [MATH]x>0[/MATH].
I have proven this inequality by letting [MATH]f(x)=\arctan x - \frac{x}{1+x^2}[/MATH] and showing that [MATH]f'(x) > 0[/MATH] for all [MATH]x>0[/MATH] and noticing that [MATH]f(0)=0[/MATH].
However I've found this interesting solution:

Letting [MATH]x=\tan \theta[/MATH] for some [MATH]\theta \in (-\pi/2,\pi/2)[/MATH] we are left with [MATH]2\theta > \sin (2\theta)[/MATH] which holds for any [MATH]\theta \in (0,\pi/2)[/MATH].

I'm unsure about some things:
1) Why did he take [MATH]\theta \in (-\pi/2,\pi/2)[/MATH] when letting [MATH]x=\tan \theta[/MATH] and why he's using the phrase "for some [MATH] \theta \in (-\pi/2,\pi/2)[/MATH]? What that phrase means? My thought is that it means something like "There exists some [MATH] \theta \in (-\pi/2,\pi/2)[/MATH] such that [MATH] x=\tan \theta[/MATH]", but I'm not sure.
If it has that meaning, is that because the tangent is a continuous function and so it assumes all the values in its range (which is [MATH]\mathbb{R}[/MATH] when [MATH]\theta \in (-\pi/2,\pi/2)[/MATH])?

2) Why did he say that the inequality holds for [MATH]\theta \in (0, \pi/2)[/MATH]? Is this because we want to show that for [MATH]x>0[/MATH] and the tangent is positive in that interval and so he cuts the negative part of the interval?

Thanks.

 

[Steven G]

Lets look at your logic more carefully.

It is fine that you defined f(x)=arctanx−x/(1+x²)

But you want to show that f(x) > 0 all x>0.

Knowing that f'(x) > 0 HAS NOTHING TO DO WITH KNOWING IF f(x)>0. Again, f'(x) > 0 says absolutely nothing about the sign of f(x).

You really need to get that down in your head! Consider y=x^3. y' = 3x^2 is always >=0. However part of f(x) = x^3 is negative and part of x^3 is positive.

You want to show that f(x)>0. If you show that the min value of f(x)>0 for x>0 then you are done. So what is the min value of f(x) for x>0??

EDIT: re-reading your last post I see that you did not just show f'(x) > 0 for x>0 BUT you also stated that f(0)=0. OK, nicely done.

 

[dunkelheit]

Thanks for your answer, actually I know that but I've written that I've noticed that [MATH]f(0)=0[/MATH] so since [MATH]f[/MATH] is increasing that should be enough to prove that [MATH]f(x)>0[/MATH] for all [MATH]x>0[/MATH], isn't it?
Actually my doubts are about the other solution I've written in "quote box".

 

[Steven G]

Prove that [MATH]\arctan x > \frac{x}{1+x^2}[/MATH] for [MATH]x>0[/MATH].
I have proven this inequality by letting [MATH]f(x)=\arctan x - \frac{x}{1+x^2}[/MATH] and showing that [MATH]f'(x) > 0[/MATH] for all [MATH]x>0[/MATH] and noticing that [MATH]f(0)=0[/MATH].
However I've found this interesting solution:

I'm unsure about some things:
1) Why did he take [MATH]\theta \in (-\pi/2,\pi/2)[/MATH] when letting [MATH]x=\tan \theta[/MATH] and why he's using the phrase "for some [MATH] \theta \in (-\pi/2,\pi/2)[/MATH]? What that phrase means? My thought is that it means something like "There exists some [MATH] \theta \in (-\pi/2,\pi/2)[/MATH] such that [MATH] x=\tan \theta[/MATH]", but I'm not sure.
If it has that meaning, is that because the tangent is a continuous function and so it assumes all the values in its range (which is [MATH]\mathbb{R}[/MATH] when [MATH]\theta \in (-\pi/2,\pi/2)[/MATH])?

2) Why did he say that the inequality holds for [MATH]\theta \in (0, \pi/2)[/MATH]? Is this because we want to show that for [MATH]x>0[/MATH] and the tangent is positive in that interval and so he cuts the negative part of the interval?

Thanks.

1) Can you please state the definition for arctan(x)? We will proceed from there.

 

[dunkelheit]

We define [MATH]\arctan x[/MATH] the inverse function of [MATH]\tan x[/MATH] by restricting (sorry if it isn't the proper term but English is not my main language) the domain of [MATH]\tan x[/MATH] to [MATH](-\pi/2,\pi/2)[/MATH]; so we have that [MATH]\arctan x[/MATH] is the angle in the interval [MATH](-\pi/2,\pi/2)[/MATH] which has [MATH]x[/MATH] has tangent.

 

[Steven G]

We define [MATH]\arctan x[/MATH] the inverse function of [MATH]\tan x[/MATH] by restricting (sorry if it isn't the proper term but English is not my main language) the domain of [MATH]\tan x[/MATH] to [MATH](-\pi/2,\pi/2)[/MATH]; so we have that [MATH]\arctan x[/MATH] is the angle in the interval [MATH](-\pi/2,\pi/2)[/MATH] which has [MATH]x[/MATH] has tangent.

Great!
Now do you see why they went from [MATH](-\pi/2,\pi/2)[/MATH]

 

[Steven G]

1) Why did he take [MATH]\theta \in (-\pi/2,\pi/2)[/MATH] when letting [MATH]x=\tan \theta[/MATH] and why he's using the phrase "for some [MATH] \theta \in (-\pi/2,\pi/2)[/MATH]? What that phrase means? My thought is that it means something like "There exists some [MATH] \theta \in (-\pi/2,\pi/2)[/MATH] such that [MATH] x=\tan \theta[/MATH]", but I'm not sure.

Yes, you are correct.