I'm currently working on some Number Theory exercises and I've hit a roadblock with this particular one. I'm a bit doubtful about my proof because the provided solution differs from mine. Not only that but it appears that the argument below leads to a sharper inequality: > instead of $\geq$. I'm afraid that I might have overlooked something, and I would really appreciate some feedback.
Show that $\pi(x) \geq \log \log x$, where $\pi(x)$ is the prime counting function.
My attempt: Let's begin by observing that the $k$-th prime, denoted as $p_k$, satisfies the inequality $p_k < 2^{2^{k}}$. Consequently, since $\log$ is strictly increasing, it follows that $\log \log p_k < k \log 2 + \log \log 2$. Expressing this in terms of the $\pi(x)$ function, $\log \log p < \pi(p) \log 2 + \log \log 2$ for any prime $p$.
Now, for an arbitrary $x \in \mathbb{N}$, define $p_0 = \operatorname{min}\{p\text{ prime};\ p > x\}$. This leads us to
\begin{align} \log \log x < \log \log p_0 < \pi(p_0) \log 2 + \log \log 2. \end{align}
Note that, by the definition of $p_0$, $\pi(x) + 1 = \pi(p_0)$. Then, \begin{align}\log \log x &< (\pi (x) + 1) \log 2 + \log \log 2 \\ &= \pi(x) \log 2 + \log 2 + \log \log 2 \\ &= \pi(x) +( - \pi(x)(1 - \log 2) + \log 2 + \log \log 2). \end{align} We know that, for $x \geq 3$, $\pi(x) \geq 2$. Therefore, the expression within the parentheses in the preceding inequality is negative for those values of $x$: \begin{align} -\pi(x) (1-\log 2)+\log 2 + \log \log 2 \leq -2(1-\log 2)+\log 2 + \log \log 2 \approx -0.28707. \end{align} Thus, $\log \log x < \pi(x)$ for every $x \geq 3$. Case $x = 2$ follows by direct verification.