Let $x$ be a real number and $\pi(x): \mathbb{R}\to\mathbb{P}$ be the function defined as
$$\pi(x)=\#\{p\le x:p \mbox{ is prime}\}$$ (that is, $\pi(x)$ outputs the number of prime numbers immediately preceding and including $x$). E.g., $\pi(2)=1$, $\pi(10) = 4$, etc.
Theorem: For $x\ge 2$, we have $\pi(x)\ge \log(\log x)$.
Proof: We first note that the result holds for $2\le x < 4$.
My comment: How can one immediately see this if not by taking the derivative of $\log(\log x))$? But since $\log$ is an increasing function, shouldn't we be able to say immediately that this result holds not just for $2\le x<4$, but for $2\le x <\infty$?
For $x\ge 4$, let $s\in\mathbb{N}$ satisfy
$$2^{2^s}\le x\le 2^{2^{s+1}}$$
[We already know from a proposition that the upper bound for $p_s$ is $2^{2^s}$]. We have that $x\ge 2^{2^s}\le p_s$, thus $\pi(x)\ge s$. By taking logarithms twice, we get:
$$x\le 2^{s+1}\implies \log x \le 2^{s+1} \log 2$$ so that $$\frac{\log\left(\frac{\log x}{\log2}\right)}{\log 2}< s+1$$
It follows that
$$\pi(x)\ge s > \frac{\log\left(\frac{\log x}{\log 2}\right)}{\log 2}-1\ge \log(\log x)$$
My comment: I'm not quite sure how to prove the last inequality.
One can show that
$$\frac{\log\left(\frac{\log x}{\log 2}\right)}{\log 2}-1=\frac{\log\left(\frac{\log x}{2\log 2}\right)}{\log 2}$$
But $2\log 2 > 1$, thus $\log\left(\frac{\log x}{2\log 2}\right)<\log \log x$.
Would appreciate some clarifications for my comments.
Well, here's a way to get a cheap lower bound.
Note that any $n≤x$ can be written uniquely as $m\times k^2$ where $m$ is square free. It follows that $$x≤ 2^{\pi(x)}\sqrt x\implies \sqrt x ≤ 2^{\pi(x)}\implies \frac {\log_2(x)}2≤\pi(x)$$ This seems better than your desired bound.