This is probably a quite simple question, but I am learning and a confirmation / explanation why is wrong would be very appreciated. Is this a valid application of Bernoulli's inequality for the first Chebyshev function?:
$$\vartheta((x+1)^2) \ge \vartheta(1+2x)$$
In other words:
$$\sum_{p \le (x+1)^2}log(p) \ge \sum_{p\le1+2x}log(p)$$
Where $p \in \Bbb P$ is a prime number.
Initially I think it is fine because $(x+1)^2 \ge 1+2x$ by Bernoulli's inequality, and that implies indirectly that the sum of the natural logarithm of the primes under those values are indeed also maintaining the same relationship, because $\vartheta$ is an increasing function so it should be correct.
Is the expression correct or is not possible such inequality? Thank you!
Yes, this is trivial; all summatory functions with non-negative summands are non-decreasing, so if $x \le y$ then $\vartheta(x) \le \vartheta(y)$. In particular, since $$1 + 2x = (1+x)^2 - x^2 \le (1+x)^2$$ for $x \ge 0$, we have $$\vartheta(1+2x) \le \vartheta((1+x)^2).$$