<s>[INST] Inequality estimate involving series Given $b_{j}>0$ and $b_{j}=b_{j+1}$ for $j=1,\ldots,n$, with $n\ge 2$ an integer and for some $\Psi>1$, does the following estimate or inequality hold?
\begin{equation}
\sum_{j=1}^n j(j-1)b_{j}|\Psi|^{j-2}~~\le~~\frac{ \left(\sum_{j=1}^n jb_j|\Psi|^{j-1}\right)^2}{\left(1+\sum_{j=1}^n b_j|\Psi|^j\right)}\nonumber
\end{equation}
Or are there constants $C>0$ and $D\ge 0$ such that the following estimate always holds for all $\Psi>1$?
\begin{equation}
\Psi^4(\Psi-1)\left(\frac{\sum_{j=1}^n j(j-1)b_j|\Psi|^{j-2}}{\left(1+\sum_{j=1}^n b_j|\Psi|^j\right)}~-~\frac{ \left(\sum_{j=1}^n j~b_j|\Psi|^{j-1}\right)^2}{\left(1+\sum_{j=1}^n b_j|\Psi|^j\right)^2}\right)
~~\le~~C\ln\left(1+\sum_{j=1}^n b_j|\Psi|^j\right)+D\nonumber
\end{equation}
Is there a neat way to prove/disprove this inequality? Note that the inequalities need only hold for some $\Psi > 1$ or $\Psi \gg 1$.
 [/INST] Substituting $x=|\Psi|$, consider a polynomial $p(x)=1+\sum_{j=1}^n b_j x^j$. Then the first inequality your asking is $$p’’(x)p(x)\le p’(x)^2$$ and the second is $$x^4(x-1)\left(\frac{p’’(x)}{p(x)}-\frac{p’(x)^2}{p(x)^2}\right)\le C\ln p(x)+D.$$
Moreover, since all $b_j=b$ are equal and $x\ne 1$, we have $p(x)=1-b +b\cdot \frac {x^{n+1}-1}{x-1}$. Therefore $p'(x)= b\cdot\frac {nx^{n+1}-(n+1)x^n+1}{(x-1)^2}$ and $p''(x)= b\cdot\frac {(n-1)nx^{n+1}-2(n-1)(n+1)x^n+n(n+1)x^{n-1}-2}{(x-1)^3}$. Hence 
$$p’(x)^2-p’’(x)p(x)=$$ $$\frac b{(x-1)^4}\left(bn x^{2n+2}+(-2bn-2b)x^{2n+1}+(bn+b)x^{2n}+((b-1)n^2+(b-1)n)x^{n+2}+((3-2b)n^2+(2b-1)n+4b-2)x^{n+1}+((b-3)n^2+(-b-1)n-2b+2) x^n+(n^2+n)x^{n-1}+(-2b+2)x+b-2)\right.$$
Since the highest coefficient of the long polynomial at the right hand side is positive, $p’’(x)p(x)-p’(x)^2$  is positive for sufficiently large $x$.
</s>