Chebyshev's bound: θ(n) < (log 4) n

We use strong induction on $n$. Base cases $n=1,2$ are immediate: $\theta(1)=0$ and $\theta(2)=\log 2 < \log 4$.

For the inductive step, suppose the bound holds for every positive integer smaller than $n$. If $n>2$ is even, $n$ is not prime, so $\theta(n)=\theta(n-1) < (\log4)(n-1) < (\log4)n$ by the induction hypothesis applied to $n-1$.

If $n=2m+1$ is odd, consider the central binomial-type coefficient $\binom{2m+1}{m}=\dfrac{(2m+1)!}{m!\,(m+1)!}$. It appears twice among the $2^{2m+1}$ terms of the binomial expansion of $(1+1)^{2m+1}$ (once as $\binom{2m+1}{m}$, once as $\binom{2m+1}{m+1}$, and these are equal), so $2\binom{2m+1}{m} \le 2^{2m+1}$, giving $\binom{2m+1}{m} \le 4^m$.

Every prime $p$ with $m+1 < p \le 2m+1$ divides the numerator $(2m+1)!$ but divides neither $m!$ nor $(m+1)!$ (since $p>m+1$), so $p$ divides $\binom{2m+1}{m}$. Hence the product of all such primes divides $\binom{2m+1}{m}$, so $\prod_{m+1<p\le 2m+1} p \le \binom{2m+1}{m} \le 4^m$, i.e. $\theta(2m+1)-\theta(m+1) \le (\log 4)\,m$.

By the induction hypothesis applied to $m+1<n$, $\theta(m+1) < (\log4)(m+1)$. Adding, $\theta(n)=\theta(2m+1) < (\log4)(m+1) + (\log4)m = (\log4)(2m+1) = (\log4)n$, completing the induction.