An Angle Bisector and the two sides...

[Labib]

In the triangle $\triangle ABC$, let $AU$ be the angle bisector of the angle $ \angle BAC$ with $U$ lying on the side $BC$.
Prove that,
$AU \leq \frac{1}{2}(AB+AC).$

[Labib]

Here's how I got going...

If I can prove, $4{AU}^2 \leq (AB+AC)^2$, it would be enough.
Let $AU=p$, $BC=a$, $AB=c$, $AC=b$, $BU=m$, $CU=n$.
With the help of Stewart's theorem, I can deduce-
${p}^2=bc-mn$
$\Rightarrow 4p^2=4bc-4mn$...............(i)
Again,
$(b+c)^2=b^2+c^2+2bc$...........(ii)
As, $b^2+c^2 \geq 2bc$
$\Rightarrow b^2+c^2+2bc > 4bc-4mn$
So from (i) and (ii), we get-
$(b+c)^2 > 4p^2$
$\Rightarrow 4AU^2 < (AB+AC)^2$
$\Rightarrow 2AU < (AB+AC)$
$\Rightarrow AU < \frac{1}{2}(AB+AC)$

Then where's the equality part??
Where's the problem??

[sourav das]

Step 1: Prove that $AU^2 = bc(1 - \frac{a^2}{(b+c)^2}) $ using Stewart's theorem and ratio theorem of angle bisector. Next substitute $a= x + y, b= y + z, c= z + x$ and find out that
$AU^2 = \frac {4y(x+y)(y+z)(x+y+z)}{(x+2y+z)^2}$. Now $(x + y) + ( y + z) \geq 2 \sqrt{(x+y)(y+z)}$ and
$( x + y + z) + y \geq 2 \sqrt{y(x+y+z)}$
And it will lead you to,
$\frac {b+c}{2} = \frac {x+2y+z}{2} \geq AU$

Stewart's theorem :http://en.wikipedia.org/wiki/Stewart%27s_theorem

[Labib]

Sourav, please check my solution for any possible mistakes!! :'(

[*Mahi*]

I was about to post the same solution Sourav Das posted, so I'd rather not waste my time writing the same solution again.
@Labib:
You said $b^2+c^2>2bc$
But when it's $b=c$?
And Stewert's Theorem: http://en.wikipedia.org/wiki/Stewart%27s_theorem

[Labib]

Mahi...
It was a typing mistake of mine...
I've edited it...
Now... If you could re-check...

[*Mahi*]

Shit! I should have seen this earlier! The equality case never occurs other than degenerate cases, that's why you never found that part. I'm using Sourav's solution for this fact.

Step 1: Prove that $AU^2 = bc(1 - \frac{a^2}{(b+c)^2}) $ using Stewart's theorem and ratio theorem of angle bisector. Next substitute $a= x + y, b= y + z, c= z + x$ and find out that
$AU^2 = \frac {4y(x+y)(y+z)(x+y+z)}{(x+2y+z)^2}$. Now $(x + y) + ( y + z) \geq 2 \sqrt{(x+y)(y+z)}$ and
$( x + y + z) + y \geq 2 \sqrt{y(x+y+z)}$
And it will lead you to,
$\frac {b+c}{2} = \frac {x+2y+z}{2} \geq AU$

But the equality part will only hold if and only if $(x + y) + ( y + z) = 2 \sqrt{(x+y)(y+z)}$ and $( x + y + z) + y = 2 \sqrt{y(x+y+z)}$ happens together, that is both $x + y= y + z$ and $ x + y + z= y$ happens together, i.e. the degenerate case, and so ,that solves it, doesn't it?

Again then in your solution, $4mn$ will be $0$, so the equality will hold...$\text{sigh}$

[bristy1588]

Now, the only problem is that the original problem was:
2AU>= AB+BC.
What modification is needed to solve this?

[bristy1588]

Now, the only problem is that the original problem was:
2AU>= AB+BC.
What modification is needed to solve this?

[*Mahi*]

WTH ?! $AU \geq \frac 1 2 (AB+BC)??$ It's not possible.
Try the simplest counter-example, a isosceles triangle.