$2^x + 3^x - 4^x + 6^x - 9^x ≤ 1$ $\forall x \in R$ How can i prove that $2^x + 3^x - 4^x + 6^x - 9^x ≤ 1$ $\forall x \in R$.
I tried $log(2^x + 3^x - 4^x + 6^x - 9^x) = log (1296^x) = x log(1296)$ i don't know if im correct  here i stuck some help please

Note that: 
$9^x-6^x+4^x-3^x-2^x+1$ 
$= 9^x-2 \cdot 6^x + 4^x + 6^x-3^x-2^x+1$
$= \left[(3^x)^2-2(3^x)(2^x) + (2^x)^2\right] + \left[(3^x)(2^x)-3^x-2^x+1\right]$
$= (3^x-2^x)^2 + (3^x-1)(2^x-1)$
$\ge 0$
since $3^x-1$ and $2^x-1$ have the same sign. 
Therefore, $2^x+3^x-4^x+6^x-9^x \le 1$
