Theorem: Smith Numbers are Infinite in Number/Lemma

Let $\map S m$ denote the sum of the digits of a positive integer $m$.
Let $\map {S_p} m$ denote the sum of the digits of the prime decomposition of $m$.
Let $\map N m$ denote the number of digits in $m$.
Suppose $m = p_1 p_2 \dots p_r$, where $p_i$ are prime numbers, not necessarily distinct.
Then $\map {S_p} m < 9 \map N m - 0.54 r$.

Proof:
Let $b_i = \map N {p_i} - 1, i = 1, 2, \dots, r$.
Let $b = b_1 + b_2 + \dots + b_r$.
Since a prime cannot be a multiple of $9$, $\map S {p_i}$ cannot be a multiple of $9$ either.
Hence:
:$\map S {p_i} \le 9 \map N {p_i} - 1 = 9 b_i + 8$
We partition the $p_i$ into $9$ disjoint classes by means of the following:
Define $c_i = \map S {p_i} - 9 b_i$.
By the above, $c_i \le 8$ and $c_i \ne 0$.
Let $n_0$ be the number of $i$ such that $c_i$ is (strictly) negative, and $n_j$ be the number of $i$ such that $c_i = j$ for $1 \le j \le 8$.
Then:
{{begin-eqn}}
{{eqn | l = \map {S_p} m
      | r = \sum_{i \mathop = 1}^r \map S {p_i}
}}
{{eqn | r = \sum_{i \mathop = 1}^r \paren {9 b_i + c_i}
}}
{{eqn | r = 9 b + \sum_{j \mathop = 1}^8 j n_j + \sum c_i
}}
{{end-eqn}}
where the last sum is over the integers $i$ where $c_i$ is (strictly) negative.
There are $n_0$ of these integers, so we also have:
{{begin-eqn}}
{{eqn | l = \sum c_i
      | o = \le
      | r = -n_0
}}
{{eqn | ll= \leadsto
      | l = \map {S_p} m
      | o = \le
      | r = 9 b + \sum_{j \mathop = 1}^8 j n_j - n_0
      | n = 1
}}
{{end-eqn}}
We have $\map S {p_i} = 9 b_i + c_i$.
For $c_i < 0$ we will use the inequality:
:$p_i > 10^{\map N {p_i} - 1} = 10^{b_i}$
For $1 \le c_i \le 8$ we would have:
:$p_i \ge \paren {c_i + 1} \times 10^{b_i} - 1$
since this number on the right is the smallest integer $N$ with $\map S N = 9 b_i + c_i$.
We also have:
:$\paren {c_i + 1} \times 10^{b_i} - 1 = \paren {c_i + 0.9} \times 10^{b_i} + 10^{b_i - 1} - 1 \ge \paren {c_i + 0.9} \times 10^{b_i}$
for $b_i > 0$ (i.e. when $p_i \ne 2, 3, 5, 7$).
Finally for $b_i = 0$, we have:
:$p_i = \map S {p_i} = c_i = c_i 10^{b_i}$
It follows that:
{{begin-eqn}}
{{eqn | l = m
      | r = p_1 p_2 \dots p_r
}}
{{eqn | o = \ge
      | r = \paren {1.9}^{n_1} \paren 2^{n_2} \paren 3^{n_3} \paren {4.9}^{n_4} \paren 5^{n_5} \paren {6.9}^{n_6} \paren 7^{n_7} \paren {8.9}^{n_8} \times 10^b
}}
{{end-eqn}}
Write $m = a \times 10^{\map N m - 1}$ where $1 \le a < 10$.
By taking logarithm (base $10$) on both sides, we get:
{{begin-eqn}}
{{eqn | l = \log a + \map N m - 1
      | o = \ge
      | r = n_1 \log 1.9 + n_2 \log 2 + \dots + n_8 \log 8.9 + b
}}
{{eqn | ll= \leadsto
      | l = 9 \map N m
      | o = \ge
      | r = 9 b + n_1 \paren {9 \log 1.9} + n_2 \paren {9 \log 2} + \dots + n_8 \paren {9 \log 8.9}
}}
{{eqn | o = \ge
      | r = 9 b + n_1 \paren {1 + 0.54} + n_2 \paren {2 + 0.54} + \dots + n_8 \paren {8 + 0.54}
      | c = by manual checking
}}
{{eqn | r = 9 b + \sum_{j \mathop = 1}^8 j n_j + 0.54 \paren {n_1 + n_2 + \dots + n_8}
}}
{{eqn | ll= \leadsto
      | l = 9 b + \sum_{j \mathop = 1}^8 j n_j
      | o = \le
      | r = 9 \map N m - 0.54 \paren {n_1 + n_2 + \dots + n_8}
      | n = 2
}}
{{end-eqn}}
Combining $(1)$ and $(2)$, we have:
{{begin-eqn}}
{{eqn | l = \map {S_p} m
      | o = \le
      | r = 9 b + \sum_{j \mathop = 1}^8 j n_j - n_0
      | c = from $(1)$
}}
{{eqn | o = \le
      | r = 9 \map N m - 0.54 \paren {n_0 + n_1 + \dots + n_8} - 0.46 n_0
      | c = from $(2)$
}}
{{eqn | o = \le
      | r = 9 \map N m - 0.54 r
}}
{{end-eqn}}
{{qed}}