Prove this inequality $\frac{n}{3^n-2^n}<\left(\frac{2}{5}\right)^{n-1}$ Let $n\ge 3$. How to prove show that
$$\dfrac{n}{3^n-2^n}<\left(\dfrac{2}{5}\right)^{n-1}\tag1$$
or 
$$3^n\cdot 2^{n-1}-2^{2n-1}-5^{n-1}\cdot n>0,n\ge 3\tag2$$
It seem right:see wolfram
when I do this problem found it:How prove this $\sum_{k=1}^{n}\frac{k}{3^k-2^k}<\frac{5}{3}$
if this inequality (1) has prove it,then
$$\sum_{k=1}^{n}\dfrac{k}{3^k-2^k}<1+\dfrac{2}{5}+\sum_{k=3}^{+\infty}\dfrac{2^{k-1}}{5^{k-1}}=\dfrac{5}{3}$$
Question: How to prove inequality (1) or (2)

$(2)$ can be proved by induction. $(2)$ is equivalent to
$$3^n\cdot 2^{n-1}>2^{2n-1}+5^{n-1}\cdot n\quad \quad (3)$$
It is easy to check the cases $n=3,4,5,6$.
Suppose $(3)$ is true for $n=k$. Then
$$3^{k+1}\cdot 2^k = 6(3^k\cdot 2^{k-1})>6(2^{2k-1}+5^{k-1}\cdot k)=\frac{3}{2}\cdot 2^{2k+1}+\frac{6}{5}\cdot 5^{k}\cdot k > 2^{2k+1}+\frac{6}{5}\cdot 5^{k}\cdot k$$
Provided $k>5$, $\frac{6}{5}k>k+1$, so the statement holds for all $n>5$ by induction.
