> Now the mean square of something that deviates around an average, as you know, is always greater than the square of the mean; so the kinetic energy integral...
No, I don't know, and you lost me. I don't think it matters to me anyway.
Since that’s still not precise, let’s compare the square of the mean to the mean square for two numbers a and b.
The square of the mean is ((a+b)/2)^2=(a^2+2ab+b^2)/4
The mean of the squares is (a^2+b^2)/2
Feynman’s claim is that the second is always bigger if the numbers deviate around an average (a and b aren’t equal).
So let’s subtract the first from the second. We get (a^2-2ab+b^2)/4. The numerator is equivalent to (a-b)^2. Since a square of a real can’t be negative, when a and b are unequal the mean of the squares is always larger.
The unaveraged variance,
∑_i (x_i - X)^2
Now, we can expand the square
0 ≤ ∑_i (x_i - X)^2 = ∑_i x_i^2 - 2 X ∑_i x_i + X^2 ∑_i 1
≤ N * (mean of the squares) - 2 N X^2 + X^2 N
≤ N * (mean of the squares) - N * (square of the mean)
square of the mean ≤ mean of the squares