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 Post subject: SIMPLE INEQUALITY
PostPosted: Mon, 29 Mar 2010 15:55:12 UTC 
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PROVE FOLLOWING INEQUALITY 1) SINA+SINB+SINC>=3/2 IF A+B+C=PIE


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 Post subject: Re: SIMPLE INEQUALITY
PostPosted: Mon, 29 Mar 2010 16:08:28 UTC 
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anmol singh wrote:
PROVE FOLLOWING INEQUALITY 1) SINA+SINB+SINC>=3/2 IF A+B+C=PIE


You can't. SINA is the SINA Corporation (USA), a NASDAQ listed company, and there is nothing called SINB as far as I know. SINC is presumably the sinc(-) function? How can you add a company, (whatever), and a function to be at least 3/2? And what can you add to physically equal to "PIE", something that you eat as dessert?

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject:
PostPosted: Mon, 29 Mar 2010 18:09:20 UTC 
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You're wrong again, outermeasure.
SINA, SINB and SINC are 3 categories of sins, and appear on the door
of Father Murphy's confessional booth.
SINA are the lighter sins; penance =< 1 Hoooly Rosary
SINB are more serious sins; 1 Hoooly Rosary < penance <= 2 Hoooly Rosaries
SINC are baaaadddddd sins (like going to Strip Clubs): 3 or more Hoooly Rosaries
(may be negociated down, proportional to amount dropped in collection basket).

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 Post subject: Re: SIMPLE INEQUALITY
PostPosted: Mon, 29 Mar 2010 18:25:18 UTC 
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anmol singh wrote:
PROVE FOLLOWING INEQUALITY 1) SINA+SINB+SINC>=3/2 IF A+B+C=PIE

I think this problem has appeared on the board before. Can anyone do a search?


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 Post subject:
PostPosted: Mon, 29 Mar 2010 19:18:22 UTC 
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This topic is hilarious, so I think I'll leave it open.

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 Post subject:
PostPosted: Mon, 29 Mar 2010 19:46:00 UTC 
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Shadow wrote:
This topic is hilarious, so I think I'll leave it open.

You won't think it's that hilarious once Father Murphy is done with you... :shock:

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 Post subject: Re: SIMPLE INEQUALITY
PostPosted: Tue, 30 Mar 2010 02:01:37 UTC 
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Matt wrote:
anmol singh wrote:
PROVE FOLLOWING INEQUALITY 1) SINA+SINB+SINC>=3/2 IF A+B+C=PIE

I think this problem has appeared on the board before. Can anyone do a search?


Seriously, it is not true.

Spoiler:
Degenerate triangle

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject:
PostPosted: Tue, 30 Mar 2010 05:42:21 UTC 
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How about if we assume that the triangle is nondegenerate?

Because the degenerate case is obviously not interesting.


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 Post subject:
PostPosted: Tue, 30 Mar 2010 05:57:48 UTC 
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Matt wrote:
How about if we assume that the triangle is nondegenerate?

Because the degenerate case is obviously not interesting.


I feel like that isn't going to be true either, \sin A+\sin B+\sin C where 0<A,B,C<\pi\quad A+B+C=\pi

Then if we chose the degenerate A=\pi,\; B=C=0 then we get 0<{3\over 2}. Tweaking the angles a bit shouldn't change the value enough to fix this problem since the function is continuous in all three variables, so I'll bet it doesn't have to be true even for non-degenerate triangles.

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 Post subject:
PostPosted: Tue, 30 Mar 2010 06:13:30 UTC 
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Maybe it is just one of those equations/inequalitites, where the letters itself are the actual unknowns :!: ...


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 Post subject:
PostPosted: Tue, 30 Mar 2010 06:22:24 UTC 
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ex falso quodlibet wrote:
Maybe it is just one of those equations/inequalitites, where the letters itself are the actual unknowns :!: ...


Maybe. We should start a mob together to get anmol to clarify his problem.

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 Post subject:
PostPosted: Tue, 30 Mar 2010 15:04:34 UTC 
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On the other hand, if it is A+B+C=pi/2, then we have \sin(A)+\sin(B)+\sin(C)\leq 3\sin((A+B+C)/3)=\frac{3}{2}.

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject:
PostPosted: Fri, 9 Apr 2010 00:47:16 UTC 
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http://www.sosmath.com/CBB/viewtopic.ph ... ht=#190924
Does it help? there is a whole discussion about a similar problem here...


I'm so hungry...I'm going to add some sines and make some pies this way :o
it's a wonderful idea

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\sqrt{\frac{1}{\mu_0 \epsilon_0}}=c


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