Mathematical proof that RMS voltage times RMS current gives mean power
Ohm's law
$$
1: V(t) = I(t)R
$$

Instantaneous power dissipation is product of voltage and current
$$
2: P(t) = V(t)I(t)\\
$$

Substitute 1 into 2 to get instantaneous power through a resistor in terms of voltage or current:
$$
3: P(t) = I^2(t)R = \frac{V^2(t)}{R}\\
$$

Average power is definitionally the integral of instantaneous power over a period, divided by that period. Substitute 3 into that to get average power in terms of voltage and current.
$$
4: P_{avg}=\frac{\int_0^T{P(t)dt}}{T}=\frac{R\int_0^T{I^2(t)dt}}{T}=\frac{\int_0^T{V^2(t)dt}}{RT}\\
$$

Definition of RMS current
$$
5: I_{RMS}=\sqrt{\frac{\int_0^T{I^2(t)dt}}{T}}\\
$$
Square both sides
$$
6: I_{RMS}^2 =\frac{\int_0^T{I^2(t)dt}}{T}\\
$$
Multiply by R to find equation 4 for average power
$$
7: I_{RMS}^2R =\frac{R\int_0^T{I^2(t)dt}}{T}=P_{avg}\\
$$
Definition of RMS voltage
$$
8: V_{RMS}=\sqrt{\frac{\int_0^T{V^2(t)dt}}{T}}\\
$$
Square both sides
$$
9: V_{RMS}^2=\frac{\int_0^T{V^2(t)dt}}{T}\\
$$
Divide by R to find equation 4 for average power
$$
10: \frac{V_{RMS}^2}{R}=\frac{\int_0^T{V^2(t)dt}}{RT}=P_{avg}\\
$$
Multiply expressions 7 and 10 for average power
$$
11: P_{avg}^2=V_{RMS}^2I_{RMS}^2\\
$$
Square root of both sides
$$
12: P_{avg} = V_{RMS}I_{RMS}\\
$$
Q.E.D.