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This is actually fairly straightforward, and reduces to the fact that

$ \sum_{p \text{prime}} \frac 1p = \log \log x + C + o(1)$

for some constant $C$. To see how to apply this to the original problem, let $p$ denote the largest prime divisor of $n$, and write $n = pz$. Then $p \ge \sqrt{n}$ if and only if $z \le p$. Thus the number of such $n \le x$ is

$\sum_{p \le x} \sum_{z \le \min(p,x/p)} 1$ which breaks up into a main term

$\sum_{p \ge \sqrt{x}} \lfloor \frac xp \rfloor$ plus a smaller term $\sum_{p \le \sqrt{x}} p$. Using Chebyshev's theorem (weaker than the PNT) that there is a constant $C> 0$ such that $\pi(x) p \le C \sqrt{x} \pi(\sqrt x/\log x$, it's easy (using partial summation) to show that = o(x)$, which is therefore negligible compared with the "smaller first term" is $O(x/\log x)$. The main term is taken care of by the equation I cited at the beginning, using the fact that $\pi(n) = o(n)$ and finally that

$\log \log x - \log \log \sqrt x = \log 2$.
show/hide this revision's text 3 Filled in the "negligible" details.

This is actually fairly straightforward, and reduces to the fact that

$ \sum_{p \text{prime}} \frac 1p = \log \log x + C + o(1)$

for some constant $C$. To see how to apply this note that to the original problem, let $n$ can have at most one p$ denote the largest prime divisor of $> n$, and write $n = pz$. Then $p \sqrt{n}$. ge \sqrt{n}$ if and only if $z \le p$. Thus the number of such $n \le x$ is

$\sum_{p\ge \sum_{p \le x} \sum_{z \le \min(p,x/p)} 1$ which breaks up into a main term

$\sum_{p \ge \sqrt{x}} \lfloor\frac{x}{p} lfloor \rfloor$,

where the sum is for frac xp \rfloor$ plus a smaller term $\sum_{p \le \sqrt{x}} p$prime.

Once you've removed Using Chebyshev's theorem (weaker than the factor of PNT) that there is a constant $p$, you are free C> 0$ such that $\pi(x) \le C x /\log x$, it's easy (using partial summation) to throw in anything you want of show that the right size"smaller term" is $O(x/\log x)$. The main term is taken care of by the equation I cited at the beginning, using the fact that $\log{2}$ comes from\pi(n) = o(n)$ and finally that

$\log \log x - \log \log \sqrt{x} sqrt x = \log 2$.
show/hide this revision's text 2 Changed big-O to little-o

This is actually fairly straightforward, and reduces to the fact that

$ \sum_{p \text{prime}} \frac 1p = \log \log x + C + O(1)$o(1)$

for some constant $C$. To see this note that $n$ can have at most one prime divisor $ > \sqrt{n}$. Thus the number such $n \le x$ is

$\sum_{p\ge \sqrt{x}} \lfloor\frac{x}{p} \rfloor$,

where the sum is for $p$ prime.

Once you've removed the factor of $p$, you are free to throw in anything you want of the right size. The $\log{2}$ comes from

$\log \log x - \log \log \sqrt{x} = \log 2$
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