Bound for Chebyshev function

Question

Consider the Chebyshev function defined as: $\psi(x)=\sum\limits_{n\leq x} \Lambda(n)$, where $\Lambda(n)=\log p$ if $n$ is a power of some prime $p$ and is equal to $0$ otherwise. Could someone please give a proper reference or a proof of the fact that $\psi(x) \geq (x-2) \log 2$ for all $x \geq 4$ ?

Answer 1

Let us consider $$I_{N}=\int_{0}^{1}x^{N}\left(1-x\right)^{N}dx.$$ Now it's easy to see that $$0<I_{N}\leq4^{-N} $$ since the maximum of the integrand is at $1/2$ and $I_{N}\in\mathbb{Q}^{+}$ since the integrand is a polynomial with integer coefficient. Furthermore the denominator that appear in the calculations of the integral are all $\leq2N+1 $. So, since holds $$e^{\psi\left(2N+1\right)}=\textrm{lcm}\left\{ 1,2,\dots,2N+1\right\}$$ we have $I_{N}e^{\psi\left(2N+1\right)}\in\mathbb{N}$ and $$I_{N}e^{\psi\left(2N+1\right)}\geq1$$ and so $$\psi\left(2N+1\right)\geq\log\left(I_{N}^{-1}\right)\geq2N\log\left(2\right) $$ then $$\psi\left(2N+1\right)\geq\left(2N+1\right)\log\left(2\right)-\log\left(2\right). $$ For the case $\psi\left(2N\right) $ we can consider $$I_{N}=\int_{0}^{1}x^{N}\left(1-x\right)^{N-1}dx $$ and using the same arguments we can conclude the proof. Note that we proved something little stronger $$\psi\left(N\right)\geq\left(N-1\right)\log\left(2\right).$$