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   (tanA)(tanB)(tanC)
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   Author  Topic: (tanA)(tanB)(tanC)  (Read 10018 times)
ThudnBlunder
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(tanA)(tanB)(tanC)  
« on: May 17th, 2008, 12:09pm »
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Prove that for any acute triangle (tanA)(tanB)(tanC) 33
« Last Edit: May 17th, 2008, 9:00pm by ThudnBlunder » IP Logged

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Re: (tanA)(tanB)(tanC)  
« Reply #1 on: May 17th, 2008, 12:40pm »
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A+B+C = pi
C = pi - (A+B)
tanC = -tan(A+B)
 
tanC = -(tanA + tanB)/(1 - tanA tanB)
 
tanA + tanB + tanC
 
= tanA + tanB - (tanA + tanB)/(1 - tanA tanB)
= {tanA(1 - tanA tanB) + tanB(1 - tanA tanB) - tanA - tanB}/(1 - tanA tanB)
= {tanA - tanA*tanA*tanB + tanB - tanA*tanB*tanB - tanA - tanB}/(1 - tanA tanB)
= -tanA*tanB(tanA + tanB)/(1 - tanA tanB)
= tanA * tanB * tanC
 
So we have  
tanA + tanB + tanC = tanA * tanB * tanC
 
or by dividing by tanA tanB tanC  
cotB cotC + cotA cotC + cotA cotB = 1.
 
By squaring both sides of S = cotA + cotB + cotC  we find  
 
  S^2 = (cotA)^2 + (cotB)^2 + (cotC)^2 + 2.  
 
Now we know that
 
 (cotA - cotB)^2 + (cotB - cotC)^2 + (cotC - cotA)^2 >= 0
 
and thus
 
 2((cotA)^2 + (cotB)^2 + (cotC)^2)  
 - 2(cotB cotC + cotA cotC + cotA cotB) >=0
 
or
 
 2(S^2 - 2) - 2 >= 0  
 2S^2 - 6 >= 0
 S^2 - 3 >= 0
 
and we find that S>= sqrt(3) (as it must be positive), which is exactly what you were looking for.
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Re: (tanA)(tanB)(tanC)  
« Reply #2 on: May 17th, 2008, 1:07pm »
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on May 17th, 2008, 12:40pm, BenVitale wrote:

So we have  
tanA + tanB + tanC = tanA * tanB * tanC
 
<snipped>
 
and we find that S>= sqrt(3) (as it must be positive), which is exactly what you were looking for.

Looks good, except 3 is not exactly what I was looking for. That would be 33  Wink
 
HINT:Once you have proved tanA + tanB + tanC = tanA * tanB * tanC you can use a well-known inequality for the rest.
« Last Edit: May 17th, 2008, 1:16pm by ThudnBlunder » IP Logged

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Re: (tanA)(tanB)(tanC)  
« Reply #3 on: May 20th, 2008, 2:28pm »
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For notational brevity, let a = tan A, b = tan B, and c = tan C.
Applying the AM-GM inequality yields
(a + b + c)/3 (abc)1/3.
 
As Ben showed, a b c = a + b + c in this case. Substituting this relation into the left hand side yields abc/3 (abc)1/3, which, after some algebra, yields abc >= 3 {3}.
 
Hence, a + b + c = abc 3 {3}.
 
« Last Edit: May 20th, 2008, 2:29pm by william wu » IP Logged


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Re: (tanA)(tanB)(tanC)  
« Reply #4 on: Aug 30th, 2008, 6:05am »
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on May 17th, 2008, 1:07pm, ThudanBlunder wrote:

HINT:well-known inequality for the rest.

 
Jensen's Inequality?
 
When A+B+C=n then  (tan A) = (tan A).
 
Let f(x)=tan x.          0<x< /2
 
The second differential of f(x) is always positive between o and /2. Hence, the function is convex. (It would have sufficed if the graph of tan x was observed, actually).
 
Jensen's inequality gives us:
 
(tan A + tan B + tan C)/3    tan ((A+B+C)/3)
 
A + B + C =  .

 
Hence Proved Smiley WOOT!
 
my first solution on this site, i think.
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