Prove that $3^{1004}+2^{2009}-3^{502}\cdot 2^{1005}\gt 2009^{182}$

Question

Prove that $3^{1004}+2^{2009}-3^{502}\cdot 2^{1005}\gt 2009^{182}$.

My try:

we have $$2^{11}\gt 2009$$

Taking power of $182$ both sides we get

$$2^{2002} \gt 2009^{182}$$

Now

$$\left(3^{1004}\right)+\left(2^{2009}\right)-\left(3^{502}\right)\left(2^{1005}\right)=2^{2002}+(127)2^{2002}+\left(3^{1004}\right)-\left(3^{502}\right)\left(2^{1005}\right)=2^{2002}+A$$

Now it suffices to prove $$A \gt 0$$ where

$$A=(127)2^{2002}+\left(3^{1004}\right)-\left(3^{502}\right)\left(2^{1005}\right)$$

any hint?

Answer 1

$\begin{align} 3^{1004}+2^{2009}-3^{502}\cdot 2^{1005} &= 3^{1004}+2^{2008}-3^{502}\cdot 2^{1005} +2^{2008}\\ &= (2^{1004}-3^{502})^2 +2^{2008}\\ &\gt 2^{2008}\\ &\gt 2009^{182} & \small\text{(per the OP argument)}\\ \end{align}$