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)
 A: 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
$$\begin{align}6^{n+1}&=6\cdot 6^n\\\\&\gt 6(4^{n}+2n\cdot 5^{n-1})\\\\&=6\cdot 4^n+12n\cdot 5^{n-1}\\\\&\ge 4\cdot 4^n+(10n+10)\cdot 5^{n-1}\\\\&=4^{n+1}+2(n+1)\cdot 5^n\qquad\blacksquare\end{align}$$
A: Some thoughts-not a full answer!
Let's try induction. For $n=3$ we readily verify that it holds: $$\frac{3}{3^3-2^3}\lt\Big(\frac25\Big)^2\Rightarrow\frac3{19}\lt\frac4{25}\Rightarrow75\lt76$$
We assume that the statement is true for $n$. We wish to show that $$\dfrac{n+1}{3^{n+1}-2^{n+1}}<\left(\dfrac{2}{5}\right)^{n}$$
But we know that $$\dfrac{n}{3^n-2^n}\lt\left(\dfrac{2}{5}\right)^{n-1}$$
We multiply this with $\Large\frac25$ to obtain $\Big(\dfrac25\Big)\dfrac{n}{3^n-2^n}\lt\left(\dfrac{2}{5}\right)^{n}$. 
Thus it suffices to show that $$\Big(\dfrac25\Big)\dfrac{n}{3^n-2^n}\gt\dfrac{n+1}{3^{n+1}-2^{n+1}}\Rightarrow\\ \\ $$ $$ \Big(\dfrac25\Big)\dfrac{3^{n+1}-2^{n+1}}{3^{n}-2^{n}}\gt\frac{n+1}{n}\Rightarrow\\ $$ $$\large\Big(\dfrac25\Big)\dfrac{\frac{3^{n+1}}{3^{n+1}}-(\frac23)^{n+1}}{\frac{3^{n}}{3^{n+1}}-\frac{2^n}{3^{n+1}}}\gt1+\frac1n\Rightarrow\\ $$ $$\large\Big(\dfrac25\Big)\dfrac{1-(\frac23)^{n+1}}{\frac{1}{3}-\frac13(\frac{2}{3})^n}\gt1+\frac1n\Rightarrow\\$$
$$ \Big(\dfrac65\Big)\dfrac{{1-(\frac23)^{n+1}}}{1-(\frac23)^n}\gt1+\frac1n$$
Now if we show that the function $$f(x)=\Big(\dfrac65\Big)\dfrac{{1-(\frac23)^{x+1}}}{1-(\frac23)^x}-1-\frac1x, x\in[1,+\infty)$$ is strictly increasing we are done since $f(1)=0$ and we will have $f(x)\gt f(1)=0$. 
This appears to be the case-here is a graph illustrating it 

Unfortunately I can't seem to go far when trying to prove that $f$ is strictly increasing.
A: $(2)$ 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.
A: $(A)\enspace$ Induction
$n=3$ : $\enspace$ That’s clear (or see $(B)$) .
Be $\enspace\displaystyle \left(\frac{6}{5}\right)^n - \left(\frac{4}{5}\right)^n - \frac{2n}{5} > 0 \enspace$ (equivalent to $(1)$)$\enspace$ for $\enspace n\geq 4\,$ .
It follows:  
$\displaystyle \left(\frac{6}{5}\right)^{n+1} - \left(\frac{4}{5}\right)^{n+1} - \frac{2(n+1)}{5} = $
$\displaystyle =\left(\frac{6}{5}\right)^n - \left(\frac{4}{5}\right)^n - \frac{2n}{5} + \frac{1}{5}\left(\left(\frac{6}{5}\right)^n + \left(\frac{4}{5}\right)^n -2\right) $ 
$\displaystyle > \frac{1}{5}\left(\left(\frac{6}{5}\right)^4 -2\right) = \frac{1}{5}(2.0736 - 2) > 0 $ 

$(B)\enspace$ Derivation
$x=3$ : $\enspace\displaystyle \frac{3}{19}=\frac{x}{3^x-2^x}|_{x=3}<\left(\frac{2}{5}\right)|_{x=3}=0.16$   
$x=4$ : $\enspace\displaystyle \frac{4}{65}=\frac{x}{3^x-2^x}|_{x=4}<\left(\frac{2}{5}\right)|_{x=4}=0.064$   
$x\in\mathbb{R}$ with $x\geq 5$ , $(1)$ multiplicated by $(3^x-2^x)\frac{2}{5}$ :
$\enspace\enspace\displaystyle \frac{2x}{5} < \left(\frac{6}{5} \right)^x - \left(\frac{4}{5}\right)^5 \leq \left(\frac{6}{5} \right)^x - \left(\frac{4}{5}\right)^x $
The right side is clear, the left side we proof by derivation.
$x=5$ : $\enspace\displaystyle 2 = \frac{2x}{5}|_{x=5} < \left(\frac{6}{5}\right)^x|_{x=5} -  \left(\frac{4}{5}\right)^5 = 2.16064 $
It’s  
$\displaystyle \frac{2}{5} = \frac{d}{dx} \frac{2x}{5} < \frac{d}{dx} (\left(\frac{6}{5}\right)^x - \left(\frac{4}{5}\right)^5 ) = \left(\frac{6}{5}\right)^x \ln \frac{6}{5} $
because of  
$\displaystyle \left(\frac{6}{5}\right)^x \geq \left(\frac{6}{5}\right)^5 = 2.48832 >  \frac{2}{5} / \ln \frac{6}{5} \approx 2.1939… \enspace$ .  
