= Solution
Write $f(n)=3^{\omega(n)}$, where $\omega$ is the <prime omega function>. If $p^a\Vert n$, its contribution to the <Dirichlet convolution> $(f*\Lambda)(n)$ is
$$
\sum_{j=1}^af(n/p^j)\log p=f(n)(a-1+1/3)\log p.
$$
Since $3(a-1+1/3)\geq a$, summing over the <prime factors> proves the <log-weighted convolution bound for three to the prime omega>, $f(n)\log n\leq3(f*\Lambda)(n)$. On $\sqrt D\leq n\leq D$, $\log n\geq\tfrac12\log D$, hence
$$
\sum_{\sqrt D\leq n\leq D}f(n)
\leq\frac6{\log D}\sum_{n\leq D}(f*\Lambda)(n)
=\frac6{\log D}\sum_{m\leq D}f(m)\psi(D/m).
$$
This is the required inequality. The standard <Chebyshev estimate> $\psi(t)\ll t$ bounds its right side by $D(\log D)^{-1}\sum_{m\leq D}f(m)/m$. To bound the latter <sum>, use <multiplicativity> and extend to all numbers with <prime factors> at most $D$:
$$
\sum_{m\leq D}\frac{f(m)}m\leq\prod_{p\leq D}\left(1+\frac3{p-1}\right).
$$
The <Mertens second theorem> states $\sum_{p\leq D}1/p=\log\log D+O(1)$. Taking <logarithms> of the product, with a summable $O(p^{-2})$ remainder, gives $3\log\log D+O(1)$. Therefore the product is $O(\log^3D)$, proving the <summatory bound for three to the prime omega> in the required range:
$$
\boxed{\sum_{\sqrt D\leq n\leq D}3^{\omega(n)}\ll D\log^2D.}
$$
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