

Title: (tanA)(tanB)(tanC) Post by ThudanBlunder on May 17^{th}, 2008, 12:09pm Prove that for any acute triangle (tanA)(tanB)(tanC) http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/geqslant.gif 3http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 

Title: Re: (tanA)(tanB)(tanC) Post by BenVitale on May 17^{th}, 2008, 12:40pm 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. 

Title: Re: (tanA)(tanB)(tanC) Post by ThudanBlunder on May 17^{th}, 2008, 1:07pm on 05/17/08 at 12:40:20, BenVitale wrote:
Looks good, except http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 is not exactly what I was looking for. That would be 3http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif3 ;) HINT:[hide]Once you have proved tanA + tanB + tanC = tanA * tanB * tanC you can use a wellknown inequality for the rest.[/hide] 

Title: Re: (tanA)(tanB)(tanC) Post by william wu on May 20^{th}, 2008, 2:28pm [hide] For notational brevity, let a = tan A, b = tan B, and c = tan C. Applying the AMGM inequality yields (a + b + c)/3 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif(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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif(abc)^{1/3}, which, after some algebra, yields abc >= 3 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{3}. Hence, a + b + c = abc http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif3 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif{3}. [/hide] 

Title: Re: (tanA)(tanB)(tanC) Post by l4z3r on Aug 30^{th}, 2008, 6:05am on 05/17/08 at 13:07:10, ThudanBlunder wrote:
[hide]Jensen's Inequality? When A+B+C=n http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif then http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/cpi.gif(tan A) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/csigma.gif(tan A). Let f(x)=tan x. 0<x<http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif /2 The second differential of f(x) is always positive between o and http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif/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 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif tan ((A+B+C)/3) A + B + C = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/pi.gif.[/hide] Hence Proved :) WOOT! my first solution on this site, i think. 

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