Prove inequality: $\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\ge2007^3.$ 
Let $a,b,c$ the sides of a triangle, $P$  its perimeter. Prove inequality:
  $$\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\ge2007^3.$$

My  attempt:
1) $P=a+b+c$. Then $\frac{P+2004a}{P-2a}=\frac{2005a+b+c}{b+c-a}$. Here $b+c-a>0 -$ triangle inequality.
2) $\sqrt[3]{xyz}\ge\frac{3}{\frac1x+\frac1y+\frac1z} \Rightarrow xyz\ge \left(\frac{3}{\frac1x+\frac1y+\frac1z}\right)^3$
 A: Let 
$$b+c-a=s,\quad c+a-b=t,\quad a+b-c=u$$
Then, we can have
$$a=\frac{t+u}{2},\quad b=\frac{s+u}{2},\quad c=\frac{s+t}{2}$$
So, using AM-GM inequality and letting $d=1003$,
$$\begin{align}&\frac{P+2004a}{P-2a}\cdot\frac{P+2004b}{P-2b}\cdot\frac{P+2004c}{P-2c}\\\\&=\left(1+d\frac ts+d\frac us\right)\left(1+d\frac st+d\frac ut\right)\left(1+d\frac su+d\frac tu\right)\\\\&=2d^3+3d^2+1+(d^3+d^2+d)\left(\frac su+\frac st+\frac tu+\frac ts+\frac us+\frac ut\right)+d^2\left(\frac{s^2}{tu}+\frac{t^2}{su}+\frac{u^2}{st}\right)\\\\&\ge 2d^3+3d^2+1+6(d^3+d^2+d)\sqrt[6]{1}+3d^2\sqrt[3]{1}\\\\&=(2d+1)^3\\\\&=2007^3\end{align}$$
A: I think it has to be $3\cdot 2007$ instead of $2007^3$. Since $a,b,c$ are sides of a triangle, there exist $x,y,z>0$ such that 
$a=y+z$, $b=z+x$ and $c=x+y$. Hence, we have to prove:
$$
\sum_{cyc}\frac{2005(x+y)+(y+z)+(z+x)}{2(x+y+z)-2(x+y)}≥3\cdot 2007\iff\\
\sum_{cyc}\frac{2006x+2006y+2z}{2z}≥3\cdot 2007\iff\\
3+1003\sum_{sym}\frac{x}{y}≥3\cdot 2007
$$
Now with AM-GM, we have $\sum_{sym}\frac{x}{y}≥6$ and thus:
$$
3+1003\sum_{sym}\frac{x}{y}≥3+6\cdot1003=3\cdot 2007
$$
Withe equality iff $x=y=z\iff a=b=c$.
