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Source. Equation (9), printed p. 87 (PDF p. 3), is attributed there to Hardy–Ramanujan. This page supplies a complete proof of the needed special case; it does not reconstruct the general theorem in the cited collected papers.

Put

$$\theta=\frac{11}{10},\qquad I=\theta\log\theta-\theta+1>0,\qquad c=\frac I4.$$

Statement. As $x\to\infty$,

$$\#\{n\le x:\Omega(n)\ge\theta\log\log x\} \le x(\log x)^{-I+o(1)} =o\!\left(\frac{x}{(\log x)^c}\right).$$

Here $\Omega$ counts prime factors with multiplicity. The same fixed $c>0$ is used in the ensuing upper bound; optimizing it is not a claim of this compilation.

Full proof

For fixed $1<z<2$, define a nonnegative multiplicative function $g$ by $g(1)=1$ and

$$g(p^a)=(z-1)z^{a-1}\qquad(a\ge1).$$

At a prime power, $1+\sum_{b=1}^a g(p^b)=z^a$. Multiplication over the prime factors proves $z^{\Omega(n)}=\sum_{d\mid n}g(d)$. Therefore

$$\begin{aligned} \sum_{n\le x}z^{\Omega(n)} &=\sum_{d\le x}g(d)\left\lfloor\frac xd\right\rfloor\\ &\le x\prod_{p\le x}\left(1+\sum_{a\ge1}\frac{g(p^a)}{p^a}\right) =x\prod_{p\le x}\left(1+\frac{z-1}{p-z}\right). \end{aligned}$$

The geometric series converge because $z<2\le p$. Moreover

$$\log\left(1+\frac{z-1}{p-z}\right) \le\frac{z-1}{p-z} =\frac{z-1}{p}+O_z(p^{-2}).$$

The reciprocal-prime estimate obtained from the prime number theorem and convergence of $\sum_p p^{-2}$ show that the product is at most $(\log x)^{z-1+o(1)}$. On the set in the statement, $z^{\Omega(n)}\ge\exp(\theta\log z\log\log x)$. Division by this threshold gives the bound

$$x(\log x)^{z-1-\theta\log z+o(1)}.$$

Set $z=\theta$. The exponent is $-I+o(1)$, and $I>0$ because $I=\int_1^\theta\log t\,dt$. Since $c<I$, the asserted little-oh estimate follows.

Scope. The source requires only a sufficiently small positive exponent in the exceptional-set bound. This elementary expansion gives one explicit admissible choice relative to the classical prime number theorem; no additional normal-order theorem is implicitly imported into the later proof.