I want to Prove the following result using induction:
Show that P$_n$: \begin{equation} 2\sum_{k=1}^n \left(\prod_{p \vert k}\left(1-\frac{1}{p}\right)\right)+2 \geq \frac{12(n-1)^2}{\pi^2}\label{eq:15} \end{equation} for all $n,k\in\mathbb{Z}_+$, where the product is over all prime factors of $k$.
Proof:
P$_1$ case: For $n = 2$ we get, \begin{equation} 2\sum_{k=1}^2 \left(\prod_{p \vert k}\left(1-\frac{1}{p}\right)\right)+2 \geq \frac{12(2-1)^2}{\pi^2} \implies 4 \geq \frac{12}{\pi^2}, \hspace{2em} \text{P$_1$ is True.} \end{equation}
P$_\alpha$ case: Suppose Eqn\eqref{eq:15} holds for some $\alpha$ s.t. \begin{equation}2\sum_{k=1}^\alpha \left(\prod_{p \vert k}\left(1-\frac{1}{p}\right)\right)+2 \geq \frac{12(\alpha-1)^2}{\pi^2} ,\hspace{2em} \forall (\alpha,k) \in \mathbb{Z}_+. \label{eq:17} \end{equation}
P$_{\alpha + 1}$ case: Now consider that it holds for some $\alpha$ + 1, \begin{equation} 2\sum_{k=1}^{\alpha + 1} \left(\prod_{p \vert k}\left(1-\frac{1}{p}\right)\right)+2 \geq \frac{12((\alpha +1)-1)^2}{\pi^2},\hspace{2em} \label{eq:18} \end{equation}
Simplifying and expanding the LHS and RHS we get, \begin{equation} 2\left(\underbrace{\prod_{p \vert 1}\left(1-\frac{1}{p}\right) + ... + \prod_{p \vert \alpha}\left(1-\frac{1}{p}\right)}_{\text{using} \ \text{P}_\alpha} + \prod_{p \vert \alpha + 1}\left(1-\frac{1}{p}\right) \right)+2 \geq \frac{12\alpha^2}{\pi^2} ,\hspace{2em} \label{eq:19} \end{equation} Therefore we get, \begin{equation} \left[2\sum_{k=1}^\alpha \left(\prod_{p \vert k}\left(1-\frac{1}{p}\right)\right) + 2\left(\prod_{p \vert \alpha + 1}\left(1-\frac{1}{p}\right)\right) \right]+2 \geq \frac{12\alpha^2}{\pi^2} ,\hspace{2em} \label{eq:20} \end{equation}
We are assuming that Eqn\eqref{eq:17} is true so, if we replace $\Phi(\alpha)$ in the above expression with the smaller number $\frac{12(\alpha-1)^2}{\pi^2}$, we produce a smaller result. So: \begin{equation}2\sum_{k=1}^\alpha \left(\prod_{p \vert k}\left(1-\frac{1}{p}\right)\right) + 2\left(\prod_{p \vert \alpha + 1}\left(1-\frac{1}{p}\right)\right) + 2 \geq\frac{12(\alpha-1)^2}{\pi^2} + 2\left(\prod_{p \vert \alpha + 1}\left(1-\frac{1}{p}\right)\right) + 2. \label{eq:21} \end{equation} Now, \begin{equation} \frac{12(\alpha-1)^2}{\pi^2} + 2\left(\prod_{p \vert \alpha + 1}\left(1-\frac{1}{p}\right)\right) + 2 = \frac{12(2-\alpha)^2}{\pi^2}. \label{eq:22} \end{equation} Thus, \begin{equation} 2\sum_{k=1}^\alpha \left(\prod_{p \vert k}\left(1-\frac{1}{p}\right)\right) + 2\left(\prod_{p \vert \alpha + 1}\left(1-\frac{1}{p}\right)\right) + 2 \geq \frac{12(2-\alpha)^2}{\pi^2} \label{eq:23} \end{equation} Since $\alpha$ $\geq$ 1, we know that $\frac{12(2-\alpha)^2}{\pi^2}$ $\geq$ 0, so \begin{equation} 2\sum_{k=1}^\alpha \left(\prod_{p \vert k}\left(1-\frac{1}{p}\right)\right) + 2\left(\prod_{p \vert \alpha + 1}\left(1-\frac{1}{p}\right)\right) + 2 \geq \frac{12(\alpha-1)^2}{\pi^2} \geq 0 \label{eq:24} \end{equation} Or, \begin{equation} 2\Phi(\alpha) + \varphi(\alpha + 1) + 2 \geq \frac{12(\alpha-1)^2}{\pi^2} \geq 0 \label{eq:25} \end{equation} Where, \begin{equation} \Phi(\alpha) = \sum_{k=1}^\alpha \left(\prod_{p \vert k}\left(1-\frac{1}{p}\right)\right) \end{equation} and \begin{equation} \varphi(\alpha + 1) = \prod_{p \vert \alpha + 1}\left(1-\frac{1}{p}\right) \end{equation} Hence, by the principle of mathematical induction, P$_n$ is true for all integers $\alpha \geq 1$.
Q.E.D
Question
I know I have made a mistake in this proof and is definitely not correct but Im not sure what I've done wrong, I think the error is in this part, \begin{equation} \frac{12(\alpha-1)^2}{\pi^2} + 2\left(\prod_{p \vert \alpha + 1}\left(1-\frac{1}{p}\right)\right) + 2 = \frac{12(2-\alpha)^2}{\pi^2} \end{equation} \begin{equation} 2\sum_{k=1}^\alpha \left(\prod_{p \vert k}\left(1-\frac{1}{p}\right)\right) + 2\left(\prod_{p \vert \alpha + 1}\left(1-\frac{1}{p}\right)\right) + 2 \geq \frac{12(2-\alpha)^2}{\pi^2} \end{equation} Since $\alpha\geq 1$, we know that $\frac{12(2-\alpha)^2}{\pi^2} \geq 0$, so, what mistake did I make and how should I fix it?
It is worse than a "mistake": it is a non-justified claim: $$\frac{12(\alpha-1)^2}{\pi^2} + 2\left(\prod_{p \vert \alpha + 1}\left(1-\frac{1}{p}\right)\right) + 2 = \frac{12(2-\alpha)^2}{\pi^2},$$ which is obviously false by irrationality of $\pi.$
Your "proof" essentially consists in this claim and is unsalvageable. Your "Result" cannot be proved by a simple induction because it would require, for every integer $\alpha\ge2$: $$2\prod_{p\mid\alpha+1}\left(1-\frac1p\right)\ge\frac{12(\alpha^2-(\alpha-1)^2)}{\pi^2},$$ which is false for every prime $\alpha+1.$
Even your "Result" itself is grossly false: test $P(4),$ or realize that the LHS of your $P(n)$ is $\le2n+2.$