Show that $(p_{k}^2+6p_{k}+8)x_{k}-(8p_{k}+24)>0$

Question

I found this exercise:

Show that $$(p_{k}^2+6p_{k}+8)x_{k}-(8p_{k}+24)>0$$ for sufficiently large $k$ where $p_{k}$ is the $k^{th}$ prime and $0<x_{k}=O\left(\frac{1}{k+1}\right)$ where $O$ is the big $O$ notation.

Answer 1

Whatever $p_k$ is (you did not specify it), the claim is false as long as $p_k$ is positive for each $k$. To see this, set $x_k = \min\{\frac{8p_k+24}{p_k^2+6p_k+8},\frac 1k\}$. Then $x_k$ decays as desired and the expression is always non-positive.

Answer 2

From $$(p_{k}^2+6p_{k}+8)x_{k}-(8p_{k}+24)>0 \iff\\ \left((p_{k}+3)^2-1\right)x_{k}-8(p_{k}+3)>0 \iff\\ (p_k+3)\left(\left((p_{k}+3)-\frac{1}{p_k+3}\right)x_{k}-8\right)>0 \iff\\$$ sinse $(p_k+3)>0$ anyway ... $$\left((p_{k}+3)-\frac{1}{p_k+3}\right)x_{k}-8>0 \iff \\ (p_{k}+3)x_{k}>8+\frac{x_{k}}{p_k+3} \iff$$ $$\color{red}{\frac{p_{k}+3}{p_{k+1}}}\cdot\color{blue}{\frac{p_{k+1}}{(k+1)\ln{(k+1)}}}\cdot\ln{(k+1)}\cdot\left(x_{k}(k+1)\right)>8+\frac{x_{k}}{p_k+3} \tag{1}$$ For the red one see this, for the blue one PNT, obviously $\ln{(k+1)}\to\infty$. From $$0<x_n=O\left(\frac{1}{k+1}\right)<\frac{M}{k+1}$$ the RHS of $(1)$ goes to $8$.

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However, the $x_{k}(k+1)$ part could lead to troubles for some $x_n=O\left(\frac{1}{k+1}\right)$. If you demand that $x_n \sim \frac{1}{k+1}$ (!!!) instead, such that $\lim\limits_{k\to\infty}x_{k}(k+1)=1$ (or any positive constant), then your claim is right, since $\ln{(k+1)}$ will outweigh $8$ in $(1)$ from some big $k$ onwards.