Inequality : arctan(x+y) <= arctanx + arctany

[Martin Lukarevski]

In a Analysis book I found this inequality which wasn't proved.

For positive x and y it holds:

arctan(x+y) <= arctanx + arctany

Who can prove it?

Martin Lukarevski

[Zdislav V. Kovarik]

In article <a1w0k2...@forum.mathforum.com>,
Martin Lukarevski <mart...@yahoo.com> wrote:
:
:
: In a Analysis book I found this inequality which wasn't proved.

:
: For positive x and y it holds:
:
: arctan(x+y) <= arctanx + arctany
:
: Who can prove it?

:
Try integrating 1/(1+(t+y)^2) - 1/(1+t^2) (which is negative for
positive t and y) by t from 0 to x.

Cheers, ZVK(Slavek).

[David C. Ullrich]

On 11 Apr 2001 09:26:48 -0400, mart...@yahoo.com (Martin Lukarevski)
wrote:

>
>
> In a Analysis book I found this inequality which wasn't proved.
>
> For positive x and y it holds:
>
> arctan(x+y) <= arctanx + arctany
>
>
> Who can prove it?

Fix y > 0. The inequality is certainly true for x = 0,
so it suffices to prove the same inequality for the
derivatives. That is, you need to show that

(*) 1/(1+(t+y)^2) <= 1/(1+t^2) ;

if you know that then integrating this inequality
from t=0 to t=x gives what you want.

I don't quite see how to prove (*)...

> Martin Lukarevski
>
>

[Peter L. Montgomery]

In article <a1w0k2...@forum.mathforum.com>
mart...@yahoo.com (Martin Lukarevski) writes:

> In a Analysis book I found this inequality which wasn't proved.

> For positive x and y it holds:

> arctan(x+y) <= arctanx + arctany

> Who can prove it?

Hint: Try fixing y. Let f(x) = arctan(x) + arctan(y) - arctan(x+y).
Show f(x) >= 0 for all x >= 0.

BTW, the difference is

xy (x + y)
f(x) = arctan ( ------------------- ) .
1 + x^2 + x*y + y^2
--
The 21st century is starting after 20 centuries complete,
but we say someone is age 21 after 21 years (plus fetus-hood) complete.
Peter-Lawren...@cwi.nl Home: San Rafael, California
Microsoft Research and CWI

[Rethnakaran P]

Hi,

I guess this can be proved algebraically using tan( a + b) expansion .
Please see the steps below.
RHS = arctanx + arctany
= arctan[Tan(arctanx + arctany)]
= arctan[(x+y)/(1-xy)]

So we have RHS - LHS = arctan[(x+y)/(1-xy)] - arctan[(x+y)]
= arctan[ {((x+y)/(1-xy)) -(x+y)}/{ 1 +
(x+y)*(x+y)/(1-xy)} ]
= arctan [ (x+y)(xy)/(1+x²+y²+xy) ]
>= 0, since arctan(ß) >= 0 for all ß >=0

Hence the result.

thanks and regards
ratna
http://ratnu.tripod.com

--
"If you tell the truth, you don't have to remember anything."
-- Mark Twain

[Peter L. Montgomery]

In article <3AD5336A...@synopsys.com>

Rethnakaran P <ra...@synopsys.com> writes:
>I guess this can be proved algebraically using tan( a + b) expansion .
>Please see the steps below.
>RHS = arctanx + arctany
> = arctan[Tan(arctanx + arctany)]
> = arctan[(x+y)/(1-xy)]
>
>So we have RHS - LHS = arctan[(x+y)/(1-xy)] - arctan[(x+y)]
> = arctan[ {((x+y)/(1-xy)) -(x+y)}/{ 1 +
>(x+y)*(x+y)/(1-xy)} ]
> = arctan [ (x+y)(xy)/(1+x²+y²+xy) ]
> >= 0, since arctan(ß) >= 0 for all ß >=0
>
>Hence the result.
>

Avoid those non-ASCII characters for squaring. Use ^2.
Also avoid beta.

Your result is correct, but there is a flaw in the derivation.
Suppose x > 1 and y > 1. Then arctan(x) + arctan(y) > pi/4 + pi/4 > pi/2.
Tan(arctan(x) + arctan(y)) will be negative (note 1 - x*y < 0).
Applying arctan will give an angle in (-pi/2, 0), not (pi/2, pi).
The equation

z = arctan(Tan(z))

fails unless -pi/2 < z < pi/2.

However your equation does hold modulo pi. And we can derive

arctan(x) + arctan(y)
== arctan(x+y) + arctan( (x+y)*x*y/(1 + x^2 + y^2 + x*y) ) mod pi.

All arctan arguments are positive, so all four arctan results
are in (0, pi/2). Each side of the == must be in (0, pi).
Since the two sides differ by a multiple of pi, they are equal.

[Martin Lukarevski]

For x>0,y>0 and 0<xy<1 the solution is elementary and the
inequality can be proved without using calculus,because then:

arctanx + arctany = arctan((x+y)/1-xy) >= arctan(x+y)
( the last inequality follows because arctan is monotonous increasing
function)

Have anyone idea for the second case when xy>1 ?

Martin Lukarevski

[Ronald Bruck]

In article <GBo00...@cwi.nl>,
"Peter L. Montgomery" <Peter-Lawren...@cwi.nl> wrote:

: Your result is correct, but there is a flaw in the derivation.

:Suppose x > 1 and y > 1. Then arctan(x) + arctan(y) > pi/4 + pi/4 > pi/2.
:Tan(arctan(x) + arctan(y)) will be negative (note 1 - x*y < 0).
:Applying arctan will give an angle in (-pi/2, 0), not (pi/2, pi).
:The equation
:
: z = arctan(Tan(z))
:
:fails unless -pi/2 < z < pi/2.
:
: However your equation does hold modulo pi. And we can derive
:
: arctan(x) + arctan(y)
: == arctan(x+y) + arctan( (x+y)*x*y/(1 + x^2 + y^2 + x*y) ) mod pi.
:
:All arctan arguments are positive, so all four arctan results
:are in (0, pi/2). Each side of the == must be in (0, pi).
:Since the two sides differ by a multiple of pi, they are equal.

I think it's worth mentioning that the inequality is valid, for x and y
POSITIVE, as a consequence of one simple fact:

(*) x \mapsto arctan(x) / x is a decreasing function on (0, pi/2).

This, in turn, is a consequence of the concavity of arctan on (0,\infty).

Then for 0 < x, y we have

arctan(x+y) arctan(x)
----------- < --------- (since x < x + y)
x + y x

arctan(x+y) arctan(y)
----------- < --------- (same reason).
x + y y

Multiply the first inequality by x, the second one by y, and add.

You can get the other cases (x < 0, y > 0; x < 0, y < 0; x = 0 or y = 0
are trivial) by exploiting the oddness of arctan.

--Ron Bruck

--
Due to University fiscal constraints, .sigs may not be exceed one
line.

[Zdislav V. Kovarik]

In article <8ov2zx...@forum.mathforum.com>,
Martin Lukarevski <mart...@yahoo.com> wrote:
:
: For x>0,y>0 and 0<xy<1 the solution is elementary and the

(At the risk of overlapping with othe people's replies:)

Avoid it altogether by turning addition into subtraction:

The formula

arctan(x+y) - arctan(y) = arctan(x / (1 + y * (x + y)))

presents no problem as long as both x>0 and y>0, and since
arctan is increasing, the right-hand side is less than

arctan(x).

Cheers, ZVK(Slavek).

[kfost...@my-deja.com]

In article <GBMzq...@cwi.nl>, Peter L. Montgomery

<Peter-Lawren...@cwi.nl> writes:
>In article <a1w0k2...@forum.mathforum.com>
>mart...@yahoo.com (Martin Lukarevski) writes:

>> In a Analysis book I found this inequality which wasn't proved.

>> For positive x and y it holds:

>> arctan(x+y) <= arctanx + arctany

>> Who can prove it?

> Hint: Try fixing y. Let

> f(x) = arctan(x) + arctan(y) - arctan(x+y).

> Show f(x) >= 0 for all x >= 0.

> BTW, the difference is

> xy (x + y)
> f(x) = arctan ( ------------------- ) .
> 1 + x^2 + x*y + y^2

This seems unnecessarily complicated. Under the given
conditions, all occurring values of arctan() are in (0, pi/2).
Say we want to determine the sense of the inequality

arctan(x+y) ? arctan(x) + arctan(y) when x > 0 and y > 0.

If arctan(x) + arctan(y) >= pi/2, the ? is trivially <.
If arctan(x) + arctan(y) < pi/2, then since the tangent
is increasing on [0, pi/2), taking the tangent of both
sides preserves the sense of the inequality, and we have

x + y ? (x+y)/(1 - xy)

Now since arctan(x) + arctan(y) < pi/2, and
arctan(x) + arctan(1/x) = pi/2 for any x > 0,
we have y < 1/x, i.e. xy < 1. So the ? is < when
x and y are both positive.

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