Consider (a+b)2; this will be the square of a negative number, and thus positive.
(a+b)2 = a2 + 2ab + b2 [algebraic identity in which all right-side terms are positive]
If we eliminate the first & last terms of the above right side, then certainly (a+b)2 ≥ 2ab.
Taking principal square roots, we have |a+b| ≥ √2•√(ab).
Dividing, we get |a+b| / √2 ≥ √(ab).
Replacing |a+b| by -(a+b) because both a & b are > 0, we have -(a+b) / √2 ≥ √(ab).
Now, note that I proved an assertion different from what you wanted--yours has the INTEGER 2 in the denominator.
I welcome feedback from all who are interested in this.