One good way for you to settle the bet by yourself it to plug in a few numbers. Let's choose u=v=1. THen clearly the left-hand side is 2 and the right-hand side must also be two. hence f=0.5. Now by your reciprocal method 1=0.5-1=-0.5, or 1= -0.5. This is evidently wrong, yet since the numbers are possible values for u, v and f we must conclude that the equation u=f-v is wrong.
So what's gone wrong? The simple answer is that only in very special circumstances does your method work (in particular see the last paragraph). Funnily enough there was a question on a Cambridge Maths exam where it was asked to carry out this analysis, so I can tell you that the equation u=f-v is valid only if f= (-v+1.73vi)/2 or f= (-v-1.73vi)/2, or put another way there are no real solutions for u, v and f that satisfy u=f-v and 1/u + 1/v =1/f. The proof of this comes from applying the correct method for rearranging the equation that I'll describe below.
Firstly when you take the reciprocal of the left-hand side you cannot do it term by term. Instead you have to say that f= 1/(1/u + 1/v). Now we apply the basic rule for adding fractions, which means that the term on the bottom must be the same for both fractions, to show that 1/u + 1/v = (u+v)/uv, so that 1/(1/u + 1/v) = uv/(u+v) = f. then uv = uf+ vf and so u(v-f)=vf or u = vf/(v-f). If we check using the values u=v=1 and f=0.5, we see that the right-hand side is now 0.5/0.5=1 as expected.
The reason that we can show that there are no real values that work with your method is by combining the two answers, u=f-v and u=vf/(v-f) and equating them. Then rearranging gives f2 +vf+v2=0, and considering this as a quadratic in f gives that the determinant is -3v2 which is always negative.
So, in conclusion, you've lost £5. Sorry.
As an exercise, you may wish to show that it is possible to have real numbers a,b,c and d such that a+b+c=d and 1/a + 1/b + 1/c =1/d, but if and only if two of a, b and c sum to zero.
*by the way, i is the square root of -1.
