<s>[INST] 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) $\sqrt[3]{xyz}\ge\frac{3}{\frac1x+\frac1y+\frac1z} \Rightarrow xyz\ge \left(\frac{3}{\frac1x+\frac1y+\frac1z}\right)^3$
 [/INST] 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 $x=y=z\iff a=b=c$.
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