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Inequality in triangle 
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#1
Nov2610, 03:12 PM

P: 58

Prove that in any triangle ABC with a sharp angle at the peak C apply inequality:(a^2+b^2)cos(αβ)<=2ab
Determine when equality occurs. I tried to solve this problem... I proved that (a^2+b^2+c^2)^2/3 >= (4S(ABC))^2, S(ABC)  area but I dont know prove that (a^2+b^2)cos(αβ)<=2ab :( thanks for your help 


#2
Nov2610, 05:09 PM

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hi harry654! welcome to pf!
(try using the X^{2} icon just above the Reply box ) by "sharp angle", i assume you mean an acute angle, less than 90°? hint: where in that triangle can you find (or construct) an angle αβ ? 


#3
Nov2710, 03:16 AM

P: 58

by "sharp angle", i assume you mean an acute angle, less than 90°? yes :)
hint: where in that triangle can you find (or construct) an angle αβ ? oh. I know that apply π(α+β)=γ. But i cant see in that triangle an angle αβ. I have tried to solve this problem already 3 days, but I always proved other inequality. 


#4
Nov2710, 03:28 AM

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Inequality in triangle
you know that αβ is in the answer, so you know there must be an αβ somewhere … where could you draw an extra line to make an angle αβ ? 


#5
Nov2710, 03:59 AM

P: 58

I can draw paralell straight line with BC. Then i draw paralell straight line with AC. And then I depict triangle ABC in axial symmetry according AB. And at the peak B(under) I have an angle αβ. Is it OK?



#6
Nov2710, 04:08 AM

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sorry, i don't understand
can you supply some extra letters (D, E, …), to make it clearer? 


#7
Nov2710, 04:19 AM

P: 58

OK.
We have triangle ABC. I draw paralell line p with BC and paralell line l with AC. p intersects l in the point D. Then angle DBA is equivalent with α. I depict triangle ABC in axial symmetry according AB. this triangle mark AEB.Then angle EBA is equivalent with β. And angle DBE is αβ 


#8
Nov2710, 04:34 AM

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oh i see!
yes, but a lot simpler would be to draw just one line … draw CD equal to CA with D on BC … then triangle DBC has two sides the same as ABC, and angle DCB is αβ (btw, "depict triangle ABC in axial symmetry according AB" is weird … write "reflect triangle ABC in AB" ) 


#9
Nov2710, 04:54 AM

P: 58

(btw, "depict triangle ABC in axial symmetry according AB" is weird … write "reflect triangle ABC in AB" )[/QUOTE]
sorry my english is not enough good I know :( draw CD equal to CA with D on BC ... When I draw D on BC so B,C,D are on one line... and it isnt triangle DBC so I am confused... sorry :( 


#11
Nov2710, 05:13 AM

P: 58

oh yes.
Now, Should I compare areas of triangles ABC and DBC, shouldnt I? 


#12
Nov2710, 05:35 AM

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no!!
(why are you so keen on areas anyway? you'll hardly ever need them, and certainly not here ) forget about triangle ABC now just use triangle DBC 


#13
Nov2710, 05:55 AM

P: 58

Ok... I use cosinus law and I have equality. But I think I must something compare because I need prove inequality



#14
Nov2710, 06:03 AM

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what equality?
(btw, we say "sine" and "cosine" ) 


#15
Nov2710, 06:14 AM

P: 58

Ok.Maybe dont understand me. When I use cosine law I get a^2+b^22abcos(αβ)=h^2
when h is DB. But I need (a^2+b^2)cos(αβ)<=2ab I now see that equality occurs when triangle is isosceles. But I try from a^2+b^22abcos(αβ)=h^2 get (a^2+b^2)cos(αβ)<=2ab. 


#16
Nov2710, 06:47 AM

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hint: eliminate ab 


#17
Nov2710, 07:28 AM

P: 58

hint: eliminate ab
I found that I am stupid... when I eliminate ab cosine law will not apply or no? 


#18
Nov2710, 08:43 AM

P: 58

apply (a²+b²)cos(αβ)>= a²+b²h²
and then I dont know aaaaa :( 


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