# inequality in triangle

by harry654
Tags: inequality, triangle
 P: 58 Prove that in any triangle ABC with a sharp angle at the peak C apply inequality:(a^2+b^2)cos(α-β)<=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(α-β)<=2ab :( thanks for your help
 PF Patron HW Helper Sci Advisor Thanks P: 25,587 hi harry654! welcome to pf! (try using the X2 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 α-β ?
 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 α-β ? oh. I know that apply π-(α+β)=γ. But i cant see in that triangle an angle α-β. I have tried to solve this problem already 3 days, but I always proved other inequality.
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P: 25,587

## inequality in triangle

 Quote by harry654 But i cant see in that triangle an angle α-β.
then make one!

you know that α-β is in the answer, so you know there must be an α-β somewhere
where could you draw an extra line to make an angle α-β ?
 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 α-β. Is it OK?
 PF Patron HW Helper Sci Advisor Thanks P: 25,587 sorry, i don't understand can you supply some extra letters (D, E, …), to make it clearer?
 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 α-β
 PF Patron HW Helper Sci Advisor Thanks P: 25,587 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 α-β (btw, "depict triangle ABC in axial symmetry according AB" is weird … write "reflect triangle ABC in AB" )
 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 :(
 PF Patron HW Helper Sci Advisor Thanks P: 25,587 sorry, i meant with D on AB
 P: 58 oh yes. Now, Should I compare areas of triangles ABC and DBC, shouldnt I?
 PF Patron HW Helper Sci Advisor Thanks P: 25,587 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
 P: 58 Ok... I use cosinus law and I have equality. But I think I must something compare because I need prove inequality
 PF Patron HW Helper Sci Advisor Thanks P: 25,587 what equality? (btw, we say "sine" and "cosine" )
 P: 58 Ok.Maybe dont understand me. When I use cosine law I get a^2+b^2-2abcos(α-β)=h^2 when h is DB. But I need (a^2+b^2)cos(α-β)<=2ab I now see that equality occurs when triangle is isosceles. But I try from a^2+b^2-2abcos(α-β)=h^2 get (a^2+b^2)cos(α-β)<=2ab.
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P: 25,587
 Quote by harry654 When I use cosine law I get a^2+b^2-2abcos(α-β)=h^2 when h is DB.
(try using the X2 icon just above the Reply box )

hint: eliminate ab
 P: 58 hint: eliminate ab I found that I am stupid... when I eliminate ab cosine law will not apply or no?
 P: 58 apply (a²+b²)cos(α-β)>= a²+b²-h² and then I dont know aaaaa :(

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