Prove this inequality for all triangles

[sharpycasio]

Homework Statement

Show that the angles a, b, c of each triangle satisfy this inequality.

$\tan \frac{a}{2}\tan \frac{b}{2} \tan \frac{c}{2} (\tan \frac{a}{2} + \tan \frac{b}{2} + \tan \frac{c}{2}) < \frac{1}{2}$

Homework Equations

The Attempt at a Solution

I used the half angle formula: $\tan \frac{\theta}{2} = \frac{1- \cos x}{\sin x}$

That gives

$$\frac{(1- \cos a)(1- \cos b)(1- \cos c)}{(\sin a)(\sin b)(\sin c)} [\frac{1- \cos a}{\sin a} + \frac{1- \cos b}{\sin b} + \frac{1- \cos c}{\sin c}] < \frac{1}{2}$$

I don't know where to go from this. Can someone give me any hints please? Thanks.

 

[SammyS]

Homework Statement

Show that the angles a, b, c of each triangle satisfy this inequality.

$\tan \frac{a}{2}\tan \frac{b}{2} \tan \frac{c}{2} (\tan \frac{a}{2} + \tan \frac{b}{2} + \tan \frac{c}{2}) < \frac{1}{2}$

Homework Equations

The Attempt at a Solution

I used the half angle formula: $\tan \frac{\theta}{2} = \frac{1- \cos x}{\sin x}$

That gives

$$\frac{(1- \cos a)(1- \cos b)(1- \cos c)}{(\sin a)(\sin b)(\sin c)} [\frac{1- \cos a}{\sin a} + \frac{1- \cos b}{\sin b} + \frac{1- \cos c}{\sin c}] < \frac{1}{2}$$

I don't know where to go from this. Can someone give me any hints please? Thanks.

One hint is that a + b + c = π .

 

[sharpycasio]

Thanks SammyS.

Since a + b + c = π we can prove that (cot A)(cot B)(cot C) = cot A + cot B + cot C
With that I did the following.

Rearrange given equation:
2[tan(a/2) + tan(b/2) + tan(c/2)] < [tan(a/2) * tan(b/2) * tan(c/2)]^-1

2[tan(a/2) + tan(b/2) + tan(c/2)] < cot(a/2) * cot(b/2) * cot (c/2)

Substitute RHS

2[tan(a/2) + tan(b/2) + tan(c/2)] < cot(a/2) + cot(b/2) + cot (c/2)

From here on, everything I try leads to no where. I tried expressing c/2 as (pi/2)-(a+b) but that makes the problem even more complicated. I honestly give up. Can someone please help me? I suck at this and I need to solve this problem for tomorrow. I'm desperate. Thanks.

 

[sharpycasio]

Can someone please please help me? I only have about two more hours to solve this stupid question.

 

[sharpycasio]

No help at all? :(

 

[SammyS]

I'm sorry, but I don't have the solution either.

My thought with a + b + c = π was simply that then

$\displaystyle c=\pi-(a+b)\ \ \text{ so that }\ \ \frac{c}{2}=\frac{\pi}{2}-\frac{a+b}{2}\,,$

thus $\displaystyle \tan\left(\frac{c}{2}\right)=\tan\left(\frac{\pi}{2}-\frac{a+b}{2}\right)=\cot\left(\frac{a+b}{2}\right)\ .$

I haven't figured out if that leads anywhere.How did you conclude the following:

Since a + b + c = π we can prove that (cot A)(cot B)(cot C) = cot A + cot B + cot C​

?

 

[sharpycasio]

I'm sorry, but I don't have the solution either.

My thought with a + b + c = π was simply that then

$\displaystyle c=\pi-(a+b)\ \ \text{ so that }\ \ \frac{c}{2}=\frac{\pi}{2}-\frac{a+b}{2}\,,$

thus $\displaystyle \tan\left(\frac{c}{2}\right)=\tan\left(\frac{\pi}{2}-\frac{a+b}{2}\right)=\cot\left(\frac{a+b}{2}\right)\ .$

I haven't figured out if that leads anywhere.


How did you conclude the following:

Since a + b + c = π we can prove that (cot A)(cot B)(cot C) = cot A + cot B + cot C​

?

http://in.answers.yahoo.com/question/index?qid=20080111225452AADKNLv

I was told that

tan(a/2)tan(b/2)+tan(a/2)tan(c/2)+tan(b/2)tan(c/2) =1 is useful.

You can prove that by taking the tan of both sides of the following

(a+b)/2 = 90 -c

Thanks for trying.

 

[SammyS]

http://in.answers.yahoo.com/question/index?qid=20080111225452AADKNLv

I was told that

tan(a/2)tan(b/2)+tan(a/2)tan(c/2)+tan(b/2)tan(c/2) =1 is useful.

You can prove that by taking the tan of both sides of the following

(a+b)/2 = 90 -c

Thanks for trying.

I looked that over & I'm quite sure there is a sign error in that derivation.

 

[ehild]

I know it is too late, but I think I found the solution just now.

As SammyS pointed out,
$$\tan(\frac{c}{2})=\cot(\frac{a+b}{2})$$

Expanding: $$\cot(\frac{a+b}{2})=\frac{1-\tan(\frac{a}{2})\tan(\frac{b}{2})}{\tan(\frac{a}{2})+\tan(\frac{b}{2})}$$

With the notations x=tan(a/2) and y=tan(b/2), (x≥0, y≥0) the original expression becomes

$$A=xy \frac{1-xy}{x+y} \left( x+y+\frac{1-xy}{x+y} \right)=xy\left(1-xy+\frac{(1-xy)^2}{(x+y)^2}\right)$$

From the relation between the arithmetic and geometric means (x+y)²≥4xy, so
$$\frac{(1-xy)^2}{(x+y)^2} ≤ \frac{(1-xy)^2}{4xy}$$

$$A≤xy\left(1-xy+\frac{(1-xy)^2}{4xy}\right)$$

denoting xy by z (z≥0):

$$A≤z\left(1-z+\frac{(1-z)^2}{4z}\right)=\frac{1}{4}(-3z^2+2z+1)$$, which is an upside-down parabola.

Check, please.

ehild