<s>[INST] 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)
 [/INST] We see that $(1)$ is equivalent to
$$6^{n}\gt 4^{n}+2n\cdot 5^{n-1}\tag3$$
Let us prove by induction that $(3)$ holds for $n\ge 3$.
We see that $(3)$ holds for $n=3,4,5$ :


*

*$6^3=216\gt 214=4^3+2\cdot 3\cdot 5^{3-1}$

*$6^4=1296\gt 1256=4^4+2\cdot 4\cdot 5^{4-1}$

*$6^5=7776\gt 7274=4^5+2\cdot 5\cdot 5^{5-1}$
Supposing that $(3)$ holds for some $n\ (\ge 5)$ gives
$$\begin{align}6^{n+1}&=6\cdot 6^n\\\\&\gt 6(4^{n}+2n\cdot 5^{n-1})\\\\&=6\cdot 4^n+12n\cdot 5^{n-1}\\\\&\ge 4\cdot 4^n+(10n+10)\cdot 5^{n-1}\\\\&=4^{n+1}+2(n+1)\cdot 5^n\qquad\blacksquare\end{align}$$
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