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 August 21st, 2013, 10:18 PM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 compare s and t Triangle ABC (area =1/4) ,with side length a,b,c and the radius of its circumscribed circle =1 given :  $t=\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}$ compare s and t
 August 22nd, 2013, 05:10 AM #2 Senior Member   Joined: Jul 2013 From: Croatia Posts: 180 Thanks: 11 Re: compare s and t Let A be area of triangle and R radius of circumscribed circle. $A=\frac{abc}{4R} \Rightarrow abc=1$ [1] $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{ab+ac+ bc}{abc}=ab+ac+bc$  RMS-AM inequality  Lemma: $a^2b^2+b^2c^2+a^2c^2 \geq a+b+c$ for a,b,c that abc=1 [2] Proof: $a^2b^2+b^2c^2 \geq 2abc \cdot b=2b$ $b^2c^2+a^2c^2 \geq 2abc \cdot c=2c$ $a^2b^2+a^2c^2 \geq 2abc \cdot a=2a$ adding we get [2] $a^2b^2+b^2c^2+a^2c^2 \geq a+b+c$ /add 2a+2b+2c $a^2b^2+b^2c^2+a^2c^2+2a+2b+2c \geq 3(a+b+c)$ $(ab+ac+bc)^2 \geq 3(a+b+c)$ $ab+ac+bc \geq \sqrt{3(a+b+c)}=3\sqrt{\frac{a+b+c}{3}}$ $t=ab+ac+bc \geq \sqrt{3(a+b+c)}=3\sqrt{\frac{a+b+c}{3}}\geq \sqrt a+\sqrt b+\sqrt c=s$
 August 22nd, 2013, 05:28 AM #3 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: compare s and t it should be s < t, or t > s
 August 22nd, 2013, 06:01 AM #4 Senior Member   Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 Re: compare s and t RMS-AM equality is only equal in the case that a=b=c. Using the sine rule and multiplying it with each other we see that $\frac {abc}{\sin( \alpha ) \sin( \beta ) \sin( \gamma )}=8$ so$\sin( \alpha ) \sin( \beta ) \sin( \gamma )=\frac {1}{8}$ If $a=b=c$ then we have a equiliteral triangle with angles of 60° $\sin( 60 ) \sin( 60 ) \sin( 60 )=\left ( \frac {\sqrt {3}}{2} \right )^3=\frac {3\sqrt {3}}{8}$ Which is a contradiction. So this means the sides of the triangle can not be equal and thus there is no equality possible. That's why we should use t>s
 August 22nd, 2013, 06:16 AM #5 Global Moderator   Joined: Dec 2006 Posts: 21,034 Thanks: 2269 How large can s/t be?
 August 24th, 2013, 04:08 AM #6 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 Re: compare s and t $A=\dfrac {abc}{4R}$ $\therefore abc=1$ $if \,\, a=b=c=1, \,\, then\,\, A\,\, =\dfrac {1\times 1\,\,sin 60^o}{2}=\dfrac {\sqrt 3}{4}>\dfrac {1}{4}=$ we get : $\dfrac {1}{a}+\dfrac {1}{b}=>2\sqrt {\dfrac {1}{ab}}=2\sqrt c-----(1)$ $\dfrac {1}{b}+\dfrac {1}{c}=>2\sqrt {\dfrac {1}{bc}}=2\sqrt a-----(2)$ $\dfrac {1}{c}+\dfrac {1}{a}=>2\sqrt {\dfrac {1}{ca}}=2\sqrt b-----(3)$ (1)+(2)+(3) and we obtain t > s
 August 24th, 2013, 05:43 AM #7 Senior Member     Joined: Feb 2010 Posts: 714 Thanks: 151 Re: compare s and t [quote="Albert.Teng"]$A=\dfrac {abc}{4R}$ $\therefore abc=1$ $if \,\, a=b=c=1, \,\, then\,\, A\,\, =\dfrac {1\times 1\,\,sin 60^o}{2}=\dfrac {\sqrt 3}{4}>\dfrac {1}{4}=$ I'm confused. If A = the area of triangle ABC, then which are you claiming is true ... A = 1/4 or A > 1/4?
 August 24th, 2013, 08:03 AM #8 Global Moderator   Joined: Dec 2006 Posts: 21,034 Thanks: 2269 The intention was presumably to show that a = b = c = 1 is not possible, as the area would not be 1/4.
August 24th, 2013, 10:21 AM   #9
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Re:

Quote:
 Originally Posted by skipjack The intention was presumably to show that a = b = c = 1 is not possible, as the area would not be 1/4.
Ahh... got it! Once that is out then the AM-GM inequality is true but with equality eliminated.

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