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
welcome to pf! 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 α-β ?
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.
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 ?
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?
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
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" )
(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 :(
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
Ok... I use cosinus law and I have equality. But I think I must something compare because I need prove inequality
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.
2ab=(a²+b²-h²)/cos(α-β) Is that OK? then a²+b²>=2ab a²+b²>=(a²+b²-h²)/cos(α-β) /*cos(α-β) (a²+b²)cos(α-β) ? (a²+b²-h²) so? I am lost...