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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,
$AU \leq \frac{1}{2}(AB+AC).$

Here's how I got going... Then where's the equality part??
Where's the problem??

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

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

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

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

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.

sourav das wrote: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...$\text{sigh}$

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

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

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