Math Is Fun Forum

Robert5

Guest

Prove

Let x,y,z > 0.

Prove that if arctanx + arctany + arctanz < pi, then x + y + z > xyz.

Thanks.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Prove

Using the identity for tan(A+B), I managed to get:

When arctanx + arctany + arctanz = pi, x + y + z - xyz = 0 ⇒ x+y+z = xyz

Not entirely sure where to go from here, as you cannot say that A < B ⇒ tanA < tanB.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Prove

Okay, I've got it.

WLOG let

, with k>0 and where A, B and C are the three angles of a right-angled triangle.

From my post above, we see that:

WLOG let

  and thus let

In this interval, tanC is increasing

Last edited by Daniel123 (2009-01-14 07:25:58)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Prove

WLOG let

No, you may not asuume WLOG that

. You may assume WLOG that

if you like, but not

.

Moreover,

would only work if

(otherwise the inequality sign is reversed). You have not shown that

.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Prove

It was stated in the initial conditions that x,y,z > 0.

Why does that initial assumption lose generality though?

Last edited by Daniel123 (2009-01-14 09:10:29)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Prove

Why does that initial assumption lose generality though?

Look at this again.

If you insist on taking WLOG. then you can only take

. But this will only lead to

and you can’t then conclude that

because

could be an obtuse angle.

Last edited by JaneFairfax (2009-01-14 10:51:56)

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Prove

A, B and C are three angles in a triangle, though? Why can't we order them arbitrarily?

Last edited by Daniel123 (2009-01-14 10:18:12)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Prove

Becauise you are taking

, not

.

Look. In order to understand what you’re doing, why not test with some particular values of

?

Suppose the three numbers are

. How does your proof fit in with that?

You would need

. Yeah? (Because you defined

, etc.)

But

are angles in a triangle.

.

Alas! Because of this, you can no longer take

! Moreover,

is negative and so

.

Sorry, Daniel. Back to square one for you.

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Prove

I see.

Thank you Jane. I don't like square one

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Prove

Your idea is good though. Maybe you just need to use it to attack the problem from another angle.

All I know is that for

, the inequality can be derived immediately from AM–GM:

So we only need consider the case

.

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Prove

I think I got it:
Let x=tanA, y=tanB, z=TanC. Then we have 0<A,B,C<pi/2 (because of arctans range). Everything now follows from the inequality:

since 0<A+B+C<pi

Since 0<A,B,C<pi/2 we have CosACosBCosC>0, so dividing by CosACosBCosC doesnt change the inequality direction, which yields:

Last edited by Kurre (2009-01-15 00:53:31)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Prove

Great work, Kurre!

Just a little point:

Then we have 0<A,B,C<pi/2 (because of arctans range).

is not imposed by the arctan range. Rather we take

to be all acute angles because

must be positive.

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Prove

Great work, Kurre!

Just a little point:

Then we have 0<A,B,C<pi/2 (because of arctans range).

is not imposed by the arctan range. Rather we take

to be all acute angles because

must be positive.

I meant that -pi/2<A,B,C<pi/2 is imposed by arctans range, then A,B,C>0 is because x,y,z>0.

But we must be careful if we just let A=arctanx, B=arctany, C=arctanz, x=tanA, y=tanB, z=tanC and consider A,B,C<pi, tanA,tanB,tanC>0 and forgets that we started with arctan and only use the positivity of x,y,z. Because now for example angles between -pi/2 and -pi would satisfies these inequalities since tan is positive in theat interval, so these problems are not completely equivalent.

Last edited by Kurre (2009-01-15 01:22:01)

Kurre

Member

Registered: 2006-07-18

Posts: 280

Re: Prove

No!

is not positive in the interval

.

I didnt say that.
Its positive in the interval ]-pi,-pi/2[.

Last edited by Kurre (2009-01-15 05:33:58)

mathsmypassion

Member

Registered: 2008-12-01

Posts: 33

Re: Prove

Sorrry Jane, but the intial problem is equivalent with:

if: 0<A,B,C<pi/2
  A+B+C<pi
then: tanA +tanB +tanC>tanAtanBtanC found by Kurre

Another way to prove that tanA +tanB +tanC>tanAtanBtanC is (not much difference compared with Kurre's solution):

tanA +tanB +tanC- tanAtanBtanC = (sinA/cosA + sinB/cosB) + tanC(1 - tanAtanB) = sin(A+B)/(cosAcosB) + tanCcos(A+B)/(cosAcosB) = sin(A+B+C)/(cosAcosBcosC) >0 when 0<A,B,C<pi/2 and A+B+C<pi

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Prove

Sorrry Jane, but the intial problem is equivalent with:

if: 0<A,B,C<pi/2
  A+B+C<pi
then: tanA +tanB +tanC>tanAtanBtanC found by Kurre

Sorry, but what is it you are trying to tell me?

crants

Guest

Re: Prove

help A+B+C=180
       tanA+tanB+tanC=x
what is tanAtanBtanC:lol: