I just wonder is this inequality true for all natural number $x$ that $$x \le\prod_{p_i\le x} \left(1+\frac {\ln x} {\ln p_i} \right)?$$
From prime factorization, we have $$x=p_{j_{1}}^{c_{j_{1}}}p_{j_{2}}^{c_{j_{2}}}\cdots p_{j_{r}}^{c_{j_{r}}}$$ for primes $p_{j_{k}}^{c_{j_{k}}}\le x$ and $c_{j_{k}}\in{\mathbb N}$ for $k=1,2,...,r$.
Then how can I proceed ?
Count all the products $$p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$$ where $p_1,p_2,\ldots,p_k$ are the prime numbers up to $x$, and $$0\le \alpha_i\le\frac{\ln x}{\ln p_i}$$ for each $i$. By unique factorisation, the number of products is equal to the number of choices for the exponents, and this is exactly $$\prod_{i=1}^k \left(1+\left\lfloor\frac {\ln x} {\ln p_i}\right\rfloor\right)\ .$$ However, any positive integer up to $x$ can be written $$x=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}$$ for some $\beta_1,\beta_2,\ldots,\beta_k$, and we have $$p_j^{\beta_j}\le x\quad\Rightarrow\quad \beta_j\le\frac{\ln x}{\ln p_j}\ .$$ Therefore the numbers we have counted include all positive integers up to $x$ (and usually, many more besides), and hence $$x\le\prod_{i=1}^k \left(1+\left\lfloor\frac {\ln x} {\ln p_i}\right\rfloor\right)\le\prod_{p_i\le x} \left(1+\frac {\ln x} {\ln p_i}\right)\ .$$
