<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] 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.
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