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Source. Lemma 4, printed p. 145 (PDF p. 3). Use $h_r$ from the factor decomposition.

Statement. For every integer $k\ge3$, real $x\ge1$ and real $Y\ge1$,

$$\#\{n\le x:h_{k(k+1)}(n)\ge Y\} \le \frac{3x}{Y^{1-2/k}}.$$

In particular, taking $Y=L(c,x)$ with $c>0$ and $x$ large gives Chen’s bound $3x/L(c(1-2/k),x)$.

Complete proof

Put $r=k(k+1)$ and

$$\mathcal H=\{m\ge1:m=h_r(m)\},\qquad S(t)=\#(\mathcal H\cap[1,t]).$$

Every prime exponent $\alpha$ in $m\in\mathcal H$ exceeds $k(k+1)$. Choose the unique $v\in\{1,\ldots,k\}$ with $v\equiv\alpha\pmod k$. Then

$$u=\frac{\alpha-v(k+1)}k$$

is a positive integer: its numerator is divisible by $k$ and positive because $\alpha>k(k+1)\ge v(k+1)$. Hence $\alpha=uk+v(k+1)$. Applying this rule to each prime represents $m=a^k b^{k+1}$. The empty product $m=1$ corresponds to $a=b=1$.

This deterministic choice maps distinct $m$ to distinct pairs $(a,b)$. For $m\le t$ we have $a\le t^{1/k}$ and $b\le t^{1/(k+1)}$, so

$$S(t)\le t^{1/k+1/(k+1)}\le t^{2/k}\qquad(t\ge1).$$

Every $n$ with $h_r(n)=m$ is divisible by $m$. Therefore

$$\#\{n\le x:h_r(n)\ge Y\} \le x\sum_{\substack{m\in\mathcal H\\m\ge Y}}\frac1m.$$

To include an atom at $m=Y$ without a boundary convention, use $1/m=\int_m^\infty t^{-2}\,dt$ and sum nonnegative integrands:

$$\sum_{\substack{m\in\mathcal H\\m\ge Y}}\frac1m =\int_Y^\infty \frac{\#\{m\in\mathcal H:Y\le m\le t\}}{t^2}\,dt \le\int_Y^\infty t^{2/k-2}\,dt =\frac{Y^{2/k-1}}{1-2/k}\le3Y^{2/k-1}.$$

This proves the statement. The construction and integral expand the two compressed steps in the source; the weak counting inequality also covers $t=1$, where the printed strict comparison cannot hold.