Let $C$ be a positive constant. Is it true that for all sufficiently large integers $n$ the inequality $$\prod_{i=1}^n (1+\frac{1}{\sqrt{p_i}})>C\sqrt{p_{n+1}}$$ holds? (Here with $p_k$ is denoted the $k$th prime number)

6$\begingroup$ Yep take logs and insert \sum_{p\leq X} 1/\sqrt{p} ~ X^{1/2} / \log{X} (by partial summation) after using \log(1+x) = x + O(x^2). $\endgroup$– alpogeAug 5 '19 at 14:46

$\begingroup$ @alpoge ...and this proves that the inequality does not hold. $\endgroup$– Fedor PetrovAug 6 '19 at 7:05

4$\begingroup$ I disagree but may be being stupid! I think the lower bound gotten is exp of \sqrt{p_n} / \log{p_n} or some such! $\endgroup$– alpogeAug 6 '19 at 7:17

$\begingroup$ @alpoge: Your comment is completely fine except that your sum is asymptotic to twice of what you indicate. $\endgroup$– GH from MOSep 5 '19 at 17:19

$\begingroup$ @GHfromMO Indeed I should know not to write $\sim$ when ignoring constants. One has to divide by the exponent 1/2 when integrating and thus the factor of two. Sorry! $\endgroup$– alpogeSep 5 '19 at 17:20
In fact, it is true. First of all denote the expression on the LHS of the inequality with $X.$ For now, fix $n$(which we will choose large enough later). Let $P$ be the set of the first $n$ primes. For a positive integer $N$ we denote by $f(N)$ the number of integers less than or equal to $N$ with squarefree parts, belonging to $P.$ Now let $A$ be a subset of $P$ and let $P(A)$ denotes the product of the elements of $A.$ The number of integers of the form $m^2A$ less than $N$ is exactly $[{\sqrt{\frac{N}{P(A)}}}],$ which is not greater than $\sqrt{\frac{N}{P(A)}}.$ Thus, $$f(N)\leq \sum_{A\subset P} \sqrt{\frac{N}{P(A)}}.$$ Note that the RHS of the inequality is exactly $$\sqrt{N}\prod_{i=1}^{n} (1+\frac{1}{\sqrt{p_i}}),$$ therefore $$X\geq \frac{f(N)}{\sqrt{N}}.$$ Now, let $M$ be a large enough integer and let $N=Mp_{n+1}.$ Let $p_{n+2},\ldots,p_{n+s}$ be all the primes greater than $p_{n+1}$ and smaller than $Mp_{n+1}.$ For each $M$ we can choose $n$ to be large enough such that $p_{n+1}>M.$ Now the numbers in $[1,N],$ which aren't counted in $f(N)$ are those divisible by $p_{n+i}(i=1,\ldots,s).$ So, $$f(N)\geq N(1\sum_{i=1}^s \frac{1}{p_{n+i}}).$$ Now, pick $n,$ such that $\log{n}>10M.$ It is well known (around the prime number theorem) that $p_n>\frac{n\log{n}}{5}$ for each $n,$ so $Mp_{n+1}>p_{n+s}>\frac{(n+s)\log{(n+s)}}{5}.$ Hence, $$n+s<\frac{5Mp_{n+1}}{\log{(n+s)}}<\frac{5Mp_{n+1}}{\log{n}}<\frac{p_{n+1}}{2}.$$ As a consequence, $$\sum_{i=1}^{s} \frac{1}{p_{n+i}}<\frac{s}{p_{n+1}}<\frac{1}{2}.$$ Therefore, $$X\geq\frac{f(Mp_{n+1})}{\sqrt{Mp_{n+1}}}\geq \frac{Mp_{n+1}}{2\sqrt{Mp_{n+1}}}\geq \frac{\sqrt{Mp_{n+1}}}{2}>C\sqrt{p_{n+1}}$$ for $M>4C^2.$

1$\begingroup$ User alpoge has already explained, in a comment below the original post, that the left hand side is much larger than the right hand side. Well, he is missing a constant, the left hand side is about the square of what he claims. Precisely, the left hand side is $\exp((2+o(1))\sqrt{n/\log n})$, while the right hand side is $(C+o(1))\sqrt{n\log n}$. $\endgroup$ Sep 5 '19 at 16:13