# Inequality in triangle

1. Nov 26, 2010

### harry654

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 :(

2. Nov 26, 2010

### tiny-tim

welcome to pf!

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 ?

3. Nov 27, 2010

### harry654

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.

4. Nov 27, 2010

### tiny-tim

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 ?

5. Nov 27, 2010

### harry654

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?

6. Nov 27, 2010

### tiny-tim

sorry, i don't understand

can you supply some extra letters (D, E, …), to make it clearer?

7. Nov 27, 2010

### harry654

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

8. Nov 27, 2010

### tiny-tim

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" )

9. Nov 27, 2010

### harry654

(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 :(

10. Nov 27, 2010

### tiny-tim

oops!

sorry, i meant with D on AB

11. Nov 27, 2010

### harry654

Re: oops!

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

12. Nov 27, 2010

### tiny-tim

no!!

(why are you so keen on areas anyway? you'll hardly ever need them, and certainly not here )

just use triangle DBC

13. Nov 27, 2010

### harry654

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

14. Nov 27, 2010

### tiny-tim

what equality?

(btw, we say "sine" and "cosine" )

15. Nov 27, 2010

### harry654

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.

16. Nov 27, 2010

### tiny-tim

(try using the X2 icon just above the Reply box )

hint: eliminate ab

17. Nov 27, 2010

### harry654

hint: eliminate ab

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

Last edited: Nov 27, 2010
18. Nov 27, 2010

### harry654

apply (a²+b²)cos(α-β)>= a²+b²-h²

and then I dont know aaaaa :(

Last edited: Nov 27, 2010
19. Nov 27, 2010

### tiny-tim

no, you've missed out a cos(α-β)

20. Nov 27, 2010

### harry654

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...