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ThudnBlunder
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 (tanA)(tanB)(tanC)   « on: May 17th, 2008, 12:09pm » Quote Modify

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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Benny
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 Re: (tanA)(tanB)(tanC)   « Reply #1 on: May 17th, 2008, 12:40pm » Quote Modify

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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ThudnBlunder
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 Re: (tanA)(tanB)(tanC)   « Reply #2 on: May 17th, 2008, 1:07pm » Quote Modify

on May 17th, 2008, 12:40pm, BenVitale wrote:
 So we have   tanA + tanB + tanC = tanA * tanB * tanC     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

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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william wu

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 Re: (tanA)(tanB)(tanC)   « Reply #3 on: May 20th, 2008, 2:28pm » Quote Modify

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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l4z3r
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 Re: (tanA)(tanB)(tanC)   « Reply #4 on: Aug 30th, 2008, 6:05am » Quote Modify

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 WOOT!

my first solution on this site, i think.
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