Set $n = 5^{2l-1}$, we have
$$E_{10}(n) = v_5(n!) = 5^{2l-2} + 5^{2l-3} + \dots + 5 + 1 = \frac{5^{2l-1}-1}{4} = \frac{n-1}{4}.$$
Since $n \equiv 2 \pmod 4$, so
$$E_9(n) = \frac{1}{2}v_3(n!) < \frac{1}{2}\left(\frac{n-2}{3} + \frac{n}{3^2} + \dots\right) = \frac{n}{4} - \frac{1}{3}.$$
Thus $E_9(n) < E_{10}(n)$.

Similarly, set $m = 3^{4l-2}$. Then we have
$$E_9(m) = \frac{1}{2}v_3(m!) = \frac{m-1}{4}$$
and since $m \equiv 4 \pmod 5$,
$$E_{10}(m) < \frac{m-4}{5} + \frac{m}{5^2} + \dots = \frac{m}{4} - \frac{4}{5}.$$
We have $E_{10}(m) < E_9(m)$