= Solution
Fix $P$ and put $e_m=h_x(mP)-m^2h_x(P)$ for $m\ge0$. Then $e_0=e_1=0$. Apply the given approximate parallelogram identity with its two arguments $mP$ and $P$. The quadratic terms cancel exactly, leaving
$$
|e_{m+1}-2e_m+e_{m-1}|\le C\qquad(m\ge1).
$$
Set $d_m=e_m-e_{m-1}$. Starting from $d_1=0$, the inequality gives $|d_m|\le(m-1)C$. Summing from one to $m$ therefore yields
$$
\boxed{|h_x(mP)-m^2h_x(P)|\le\frac{m(m-1)}2C.}
$$
Thus $C_m=m(m-1)C/2$ works independently of $P$, including $m=1$.
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