I found a horrible, ugly solution using basic calculus, and it requires manually checking 2<=m<=5.

$2a_n-3a_{n-1}=\frac{3}{2^{n+1}}\Longleftrightarrow a_n+\frac{3}{2^{n+3}}=\frac{3}{2}\left(a_{n-1}+\frac{3}{2^{n+2}}\right)$,so we have $a_n+\frac{3}{2^{n+3}}=\left(\frac{3}{2}\right)^{n-1}\left(a_1+\frac{3}{2^4}\right)=\left(\frac{3}{2}\right)^n$. Therefore we can rewrite the given inequality as follows. \[\left(\frac{3}{2}\right)^{\frac{n}{m}}\left[m-\left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}\right]<\frac{m^2-1}{m-n+1}\] $1-\left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}=1-\left[\left(\frac{2}{3}\right)^{\frac{m-1}{m}}\right]^n<n\left[1-\left(\frac{2}{3}\right)^{\frac{m-1}{m}}\right]$ $\Longleftrightarrow m-\left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}<m+n-1-n\left(\frac{2}{3}\right)^{\frac{m-1}{m}}$

kunny wrote: Therefore we can rewrite the given inequality as follows. \[\left(\frac{3}{2}\right)^{\frac{n}{m}}\left[m-\left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}\right]<\frac{m^2-1}{m-n+1}\] And from here, note that when $n=0$, we have equality. So think of it as a function in n, for $m \geq 6$, it suffices to show that this is monotonic increasing. This is a straight forward derivative check. One can then manually check when $2\leq m \leq 5$

As it was noticed by kunny,our inequality is equivalent to: $ (\frac {3}{2})^{\frac {n}{m}}(m - (\frac {2}{3})^{\frac {n(m - 1)}{m}}) < \frac {m^2 - 1}{m - n + 1}$ After that we will work with RHS: ${ \frac {m^2 - 1}{m - n + 1} = (m - 1)(\frac {n}{m - n + 1} + 1})$. Using Bernoulli inequality: ${ (\frac {n}{m - n + 1} + 1})^{\frac {m}{2n}}\geq(\frac {m}{2m - 2n + 2} + 1)$,but as we know $ m\geq n\geq 1$,thus $ \boxed{\frac {m}{2m - 2n + 2} + 1\geq\frac {3}{2}}$,so ${ \frac {m^2 - 1}{m - n + 1} = (m - 1)(\frac {n}{m - n + 1} + 1})\geq(m - 1)(\frac {3}{2})^{\frac {2n}{m}}$. So we need to prove that: $ \boxed{(\frac {2}{3})^{\frac {n}{m}}(m - (\frac {2}{3})^{\frac {n(m - 1)}{m}}) < m - 1}$,or $ (\frac {2}{3})^{\frac {n}{m}}\cdot m - (\frac {2}{3})^{n} < m - 1$,or $ ((\frac {2}{3})^{\frac {n}{m}} - 1)(m - 1) < (\frac {2}{3})^{n} - 1$. Now let $ (\frac {2}{3})^{\frac {n}{m}} = x$: $ (x - 1)m < (x - 1)(x^{m - 1} + x^{m - 2} + \dots + x + 1)$,but as we know $ x < 1$ and $ x^{m - 1} + x^{m - 2} + \dots + 1 < m$.

Erken wrote: Using Bernoulli inequality: ${ (\frac {n}{m - n + 1} + 1})^{\frac {m}{2n}}\geq(\frac {m}{2m - 2n + 2} + 1)$,but as we know $ m\geq n\geq 1$. It is wrong,because Bernoulli inequality requires ${\frac {m}{2n}}\ge 1$.

kunny wrote: \[\left(\frac{3}{2}\right)^{\frac{n}{m}}\left[m-\left(\frac{2}{3}\right)^{\frac{n(m-1)}{m}}\right]\le \frac{m^2-1}{m-n+1}\] I prove this ineq as follows: Let $(\frac{2}{3})^\frac{n}{m}=a\ge \frac{2}{3}$ thus $n=m\log_{\frac{2}{3}}a$ Now, the ineq equivalent to: $\frac{(m^2-1)a}{m+1-m\log_{\frac{2}{3}}a}>m-a^{m-1}$ Considering the law: $f(a)=LHS-RHS$ Well to known: $f'(a)=\frac{(m^2-1)(m+1-n-m\ln\frac{2}{3})}{(m-n+1)^2}+(m-1)a^{m-2}>0$ ( because $\ln\frac{2}{3}<0$) From which we get: $f(a)\ge f(\frac{2}{3})=\frac{2}{3}(m^2-1)-m+(\frac{2}{3})^{m-1}\ge 0$ for all $m\in Z_+$. Done