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Write $$ Q(x)=m(m+1)\left[x-\frac{m-1}{m+1}\right]^2-2\frac{m-1}{m+1} $$ Note that $Q(x)\ge -2\frac{m-1}{m+1}$, so $[2u+u^2]Q(x)\ge -6\frac{m-1}{m+1}u$. Also put $x=\frac{m-1-t}{m+1}$ (clearly, we are in more trouble if we go left from the parabola vertex: the value of the quadratic form is the same as at the corresponding distance on the right but $u$ is smaller). Thus, it will suffice to ensure that $$ \left(\alpha-6\frac{m-1}{m+1}\right)\left[\frac{m-1-t}{m+1}\right]^m\ge (2-t^2)\frac{m-1}{m+1} $$ Hence, we can take $\alpha=\frac{m-1}{m+1}\left(6+\max_t (2-t^2)\left[\frac{m+1}{m-1-t} \right]^m\right)$. Now observe that the RHS is a decreasing function of $m$ for a fixed $t$. Indeed, the Taylor expansion of $\dfrac{ \log(1+\frac 1m)-\log(1-\frac{1+t}m)}{\frac 1m}$ in $\frac 1m$ has all the coefficients positive. Thus we can take $m=4$. This value is nice because the equation for the critical point becomes $\frac{2t}{1-t^2}=\frac 4{3-t}$ and its root $t=1$ is obvious (of course, we should also check that it is the only positive root afterwards but it rewrites as $3t+t^2=4$). Plugging $t=1$ in, we get $[\frac 52]^4=\frac{625}{16}\le 40$. Thus we are always happy with $\alpha=46\frac {m-1}{m+1}$ units. The actual reserve is $4m^2-16m+10=4(m-4)m+10$. If $m\ge 6$, this is at least $58$, which is more than enough regardless of $m$. For $m=5$, we have $30$ and it looks like we need $30\frac 13$. However notice that if $u\ge\frac 12$, say, then $24u$ is already way above $2$, so we can safely assume that $u\le \frac 12$, which immediately frees up $8$ extra units in this case.

Write $$ Q(x)=m(m+1)\left[x-\frac{m-1}{m+1}\right]^2-2\frac{m-1}{m+1} $$ Note that $Q(x)\ge -2\frac{m-1}{m+1}$, so $[2u+u^2]Q(x)\ge -6\frac{m-1}{m+1}u$. Also put $x=\frac{m-1-t}{m+1}$ (clearly, we are in more trouble if we go left from the parabola vertex: the value of the quadratic form is the same as at the corresponding distance on the right but $u$ is smaller). Thus, it will suffice to ensure that $$ \left(\alpha-6\frac{m-1}{m+1}\right)\left[\frac{m-1-t}{m+1}\right]^m\ge (2-t^2)\frac{m-1}{m+1} $$ Hence, we can take $\alpha=\frac{m-1}{m+1}\left(6+\max_t (2-t^2)\left[\frac{m+1}{m-1-t} \right]^m\right)$. Now observe that the RHS is a decreasing function of $m$ for a fixed $t$. Indeed, the Taylor expansion of $\dfrac{ \log(1+\frac 1m)-\log(1-\frac{1+t}m)}{\frac 1m}$ in $\frac 1m$ has all the coefficients positive. Thus we can take $m=4$. This value is nice because the equation for the critical point becomes $\frac{2t}{2-t^2}=\frac 4{3-t}$ and its root $t=1$ is obvious (of course, we should also check that it is the only positive root afterwards but it rewrites as $3t+t^2=4$). Plugging $t=1$ in, we get $[\frac 52]^4=\frac{625}{16}\le 40$. Thus we are always happy with $\alpha=46\frac {m-1}{m+1}$ units. The actual reserve is $4m^2-16m+10=4(m-4)m+10$. If $m\ge 6$, this is at least $58$, which is more than enough regardless of $m$. For $m=5$, we have $30$ and it looks like we need $30\frac 13$. However notice that if $u\ge\frac 12$, say, then $24u$ is already way above $2$, so we can safely assume that $u\le \frac 12$, which immediately frees up $8$ extra units in this case.

Now we have to assess the cost of this in our units, i.e., to find the best $\beta$ such that $$ \frac{\beta}{(1+z^m)^2}\ge m\frac{m-(m+2)x^2+2x^{m+1}} {(1-x^m)^2} $$ or, equivalently $$ \beta (1-z^{m})^2\ge m (1+z^m)^2(m-(m+2)x^2+2x^{m+1} )\,. $$ Plugging in $x=0$ (in a trivial way) and $x=1$ (after taking the second derivative), we see that $\beta\ge\max(m^2,4(m+2))$. Now let us show that this maximum really works. Dividing by $(1-x)^2$, we can rewrite it as $$ \beta(1+x+\dots+x^{m-1})^2\ge m(1+2x^m+x^{2m})(m+2mx+(2m-2)x^2+(2m-4)x^3+\dots+4x^{m-1}+2x^m) $$ Notice now that the coefficients of the LHS (after opening the parentheses) first go up by $\beta$ and then, after position $m-1$, start dropping by $\beta$ at each step. On the other hand, on the RHS we cannot go up faster than by $m^2$ up to position $m-1$ and beyond that position, the largest drop is $4$ throughout the whole sequence. Thus the coefficient sequence of the LHS dominates at least up to position $m-1$ and then, once it starts going down, there may be at most one crossing. Also the total sum of coefficients for the LHS is at least that for the RHS (that's guaranteed by our checking the second derivative at $x=1$). But if we have any two polinomials like that, the left one dominates the right one on $[0,1]$, so we are done here. Since $4(m+2)\le m^2+3$ for $m\ge 5$ (the range we are aiming at), let us put $\beta=m^2+3$ always to keep the formulae neat.

Now we have to assess the cost of this in our units, i.e., to find the best $\beta$ such that $$ \frac{\beta}{(1+x^m)^2}\ge m\frac{m-(m+2)x^2+2x^{m+1}} {(1-x^m)^2} $$ or, equivalently $$ \beta (1-x^{m})^2\ge m (1+x^m)^2(m-(m+2)x^2+2x^{m+1} )\,. $$ Plugging in $x=0$ (in a trivial way) and $x=1$ (after taking the second derivative), we see that $\beta\ge\max(m^2,4(m+2))$. Now let us show that this maximum really works. Dividing by $(1-x)^2$, we can rewrite it as $$ \beta(1+x+\dots+x^{m-1})^2\ge m(1+2x^m+x^{2m})(m+2mx+(2m-2)x^2+(2m-4)x^3+\dots+4x^{m-1}+2x^m) $$ Notice now that the coefficients of the LHS (after opening the parentheses) first go up by $\beta$ and then, after position $m-1$, start dropping by $\beta$ at each step. On the other hand, on the RHS we cannot go up faster than by $m^2$ up to position $m-1$ and beyond that position, the largest drop is $4$ throughout the whole sequence. Thus the coefficient sequence of the LHS dominates at least up to position $m-1$ and then, once it starts going down, there may be at most one crossing. Also the total sum of coefficients for the LHS is at least that for the RHS (that's guaranteed by our checking the second derivative at $x=1$). But if we have any two polinomials like that, the left one dominates the right one on $[0,1]$, so we are done here. Since $4(m+2)\le m^2+3$ for $m\ge 5$ (the range we are aiming at), let us put $\beta=m^2+3$ always to keep the formulae neat.

To be continued...

Now, as promised, the special case $n=1$. Again, I'll make no clever steps, just one insolent one.

The function in this case reduces to $\log\left[(1+x^5)\frac{1-x^3}{1-x^2}\right]$. Honestly taking the second derivative, we see that we need to check that $$ \frac{20-5x^5}{(1+x^5)^2}x^3-\frac{6+3x^3}{(1-x^3)^2}x+\frac{2+2x^2}{(1-x^3)^2}\ge 0\,. $$ Now I would like to multiply by the common denominator, but it will take the computational skills of Euler to multiply 3 squares of binomials, so, I'll just reduce the first term to $x^3$ (not even $3x^3$: I want to keep my numbers small, because otherwise an arithmetic error is inavoidable somewhere and I do not want to use any CAS here). I'll spare you from the exercise of multiplying out 4 binomials once and 3 binomials twice (takes about 15 minutes) and just write the result: you need to show that the polynomial with the (infinite) integer coefficient sequence $$ [2,-6,2,9,-3,-12,6,1,3,1,-2,-2,0,1,\bar 0] $$ is non-negative on $[0,1]$.

Now I know just one human-friendly criterion for such problems: start replacing sequences by their partial sums like crazy until you either get a negative tail (failure) or a non-negative sequence (success). In order to make sure that one of the two occurs, you should factor out rational roots on $(0,1)$, of course, but you normally just start and see what happens.

We get successively $$ \begin{aligned} &[2,-4,-2,7,4,-8,-2,-1,2,3,1,-1,-1,\bar 0] \\ &[2,-2,-4,3,7,-1,-3,-4,-2,1,2,1,\bar 0] \\ &[2,0,-4,-1,6,5,2,-2,-4,-3,-1,\bar 0] \\ &[2,2,-2,-3,3,8,10,8,4,1,\bar 0] \\ &[2,4,2,-1,2,10,\bar +] \\ &[2,6,8,7,9,\bar+] \end{aligned} $$ Success!

Conclusion

Formally the question is now answered in full as originally posted. I don't expect you to like this solution (it certainly lacks in uplift) but, alas, I don't expect anyone to find a nice way to handle this problem (some steps can be done in a more elegant way, as I see now, but to retype them would be too much trouble, so I'll not do it unless somebody requests it). As usual, nothing will give me more pleasure than being proved wrong in my pessimistic predictions. :-)