# Are the degrees of freedom for Welch's test always less than the DF of the pooled test?

I am teaching a course on basics statistics, and we are doing the t-test for two independent samples with unequal variances (Welch test). In the examples I have seen, the adjusted degrees of freedom used by the Welch test are always less than or equal to $n_1+n_2-2$.

Is this always the case? Does the Welch test always reduce (or leave unchanged) the degrees of freedom of the pooled (equal variances) t-test?

And on the same subject, if the sample standard deviations are equal, do the DF's of the Welch test reduce to $n_1+n_2-2$? I looked at the formula, but the algebra got messy.

$$df'=\frac{\left(\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}\right)^2}{\frac{\left(\frac{s^2_1}{n_1}\right)^2}{n_1-1}+\frac{\left(\frac{s^2_2}{n_2}\right)^2}{n_2-1}}$$ As you can see, it's rather ugly (and in fact is approximated numerically), but it necessitates that $df'<df$. Here's a reference: Howell (2002, p. 214) states that, "$df'$ is bounded by the smaller of $n_1-1$ and $n_2-1$ at one extreme and by $n_1+n_2-2~df$ at the other".