When does equality occur in the inequality (a^2+b^2)cos(α-β)<=2ab?

[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 don't know prove that (a^2+b^2)cos(α-β)<=2ab :(
thanks for your help

 

[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 ?

 

[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 can't see in that triangle an angle α-β. I have tried to solve this problem already 3 days, but I always proved other inequality.

 

[tiny-tim]

But i can't 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 ? ​

 

[harry654]

I can draw parallel straight line with BC. Then i draw parallel 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?

 

[tiny-tim]

sorry, i don't understand

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

 

[harry654]

OK.
We have triangle ABC.
I draw parallel line p with BC and parallel 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

 

[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" )

 

[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 isn't triangle DBC so I am confused... sorry :(

 

[tiny-tim]

oops!

sorry, i meant with D on AB

 

[harry654]

oh yes.
Now, Should I compare areas of triangles ABC and DBC, shouldn't I?

 

[tiny-tim]

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

 

[harry654]

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

 

[tiny-tim]

what equality?

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

 

[harry654]

Ok.Maybe don't 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.

 

[tiny-tim]

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

 

[harry654]

hint: eliminate ab

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

 

[harry654]

apply (a²+b²)cos(α-β)>= a²+b²-h²

and then I don't know aaaaa :(

 

[tiny-tim]

apply (a²+b²)cos(α-β)>= a²+b²-h²

no, you've missed out a cos(α-β)

 

[harry654]

no, you've missed out a cos(α-β)

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...

 

[tiny-tim]

i can't see where you got lost …

you should have arrived at (a²+b²)cos²(α-β) ≤ a²+b² - h² …

carry on from there ​

 

[harry654]

you should have arrived at (a²+b²)cos²(α-β) ≤ a²+b² - h² … ( but why is there ≤ and not >=)

(a²+b²)cos²(α-β) ≤ a²+b² - h²


(a²+b²)cos(α-β) ≤ 2ab

 

[tiny-tim]

(a²+b²)cos²(α-β) ≤ a²+b² - h²


(a²+b²)cos(α-β) ≤ 2ab

sorry, I'm not following you at all

 

[harry654]

thank you for your patience
I think I never solve this problem

 

[harry654]

How do you get (a²+b²)cos²(α-β) ≤ a²+b² - h² ?

 

[harry654]

I really don't know how carry on...
I have a cosine law and then I don't know how get (a²+b²)cos²(α-β) ≤ a²+b² - h²
Please help me:(

 

[tiny-tim]

I have a cosine law …/QUOTE]

what cosine law did you use?

 

[harry654]

I use this cosines law h²=a²+b²-2abcos(α-β) h is |DB| and then I don't know how carry on

 

[tiny-tim]

I use this cosines law h²=a²+b²-2abcos(α-β) h is |DB|

ok then 2ab = (a²+b²-h²)/cos(α-β) …

substitute that into the inequality

 

[harry654]

so and now the problem begins because I don't know how :(

 

[tiny-tim]

uhh?

just write out the inequality, with 2ab replaced by (a²+b²-h²)/cos(α-β)

 

[harry654]

yes I see, but what inequality do you think ?

 

[tiny-tim]

yes I see, but what inequality do you think ?

what are you talking about?

there is only one inequality

 

[harry654]

but I can't
in (a²+b²)cos(α-β)≤2ab
substitude 2ab because I must prove that inequality and when I substitude 2ab it isn't mathematical proof

 

[tiny-tim]

do it now to see where you're going …

you can tidy it up later! ​

 

[harry654]

OK
I have (a²+b²)cos(α-β) ≤ (a²+b²-h²)/cos(α-β)
and from it I get
(a²+b²)cos²(α-β) ≤ a²+b²-h²
but how can I tidy up later?

 

[tiny-tim]

oh, forget about later!

ok, now simplify (a²+b²)cos²(α-β) ≤ a²+b²-h²

 

[harry654]

OK
(a²+b²)cos²(α-β) ≤ a²+b²-h²
(a²+b²)cos²(α-β) ≤ a²+b²-a²-b²+2abcos(α-β)
(a²+b²)cos²(α-β) ≤ 2abcos(α-β)
(a²+b²)cos(α-β) ≤ 2ab

 

[tiny-tim]

OK
(a²+b²)cos²(α-β) ≤ a²+b²-h²
(a²+b²)cos²(α-β) ≤ a²+b²-a²-b²+2abcos(α-β)
(a²+b²)cos²(α-β) ≤ 2abcos(α-β)
(a²+b²)cos(α-β) ≤ 2ab

uhh? … now you're going backwards

simplify (a²+b²)cos²(α-β) ≤ a²+b²-h²

 

[harry654]

(a²+b²)cos²(α-β) ≤ a²+b²-a²-b²+2abcos(α-β)
(a²+b²)cos²(α-β) ≤ 2abcos(α-β) OK?

 

[tiny-tim]

(a²+b²)cos²(α-β) ≤ a²+b²-a²-b²+2abcos(α-β)
(a²+b²)cos²(α-β) ≤ 2abcos(α-β) OK?

you're still going backwards, that's exactly the same as (a²+b²)cos(α-β) ≤ 2ab

 

[harry654]

and how can I simplify (a²+b²)cos²(α-β) ≤ a²+b²-h² ? I can only as (a²+b²)cos²(α-β) ≤ a²+b²-a²-b²+2abcos(α-β) :(

 

[tiny-tim]

well, for a start, you could try putting all similar terms on the same side

 

[harry654]

similar terms on the same side? uff

 

[harry654]

I don't know how carry on again :(

 

[harry654]

could you write me an inequality at which should I arrive?

 

[harry654]

I tried this
(a²+b²)cos²(α-β) ≤ a²+b²-h²
(a²+b²)cos²(α-β)-(a²+b²)≤ -h²
but how carry on... I am desperate:(

 

[tiny-tim]

good morning!

(a²+b²)cos²(α-β)-(a²+b²)≤ -h²

yes, that's a standard way of simplifying an equation …

putting all the similar things on one side!

ok, now simplify (a²+b²)cos²(α-β)-(a²+b²) ​

 

[harry654]

good morning
(a²+b²)cos²(α-β)-(a²+b²)=
=(a²+b²)(cos²(α-β)-1)=
=(a²+b²)(cos(α-β)-1)(cos(α-β)+1)
OK?

 

[tiny-tim]

good morning
(a²+b²)cos²(α-β)-(a²+b²)=
=(a²+b²)(cos²(α-β)-1)=
=(a²+b²)(cos(α-β)-1)(cos(α-β)+1)
OK?

oh good grief!

no wonder you've been having difficulty with this question

take the morning off and become familiar with using your https://www.physicsforums.com/library.php?do=view_item&itemid=18" …

in particular cos² + sin² = 1 ​