A geometric look at $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$? Is there a geometric way of looking at the relationship between the positive real numbers $a$, $b$ and $c$ if $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}$?
 A: You find this kind of “addition” happening frequently. For instance, if you have lenses of focal length $a$ and $b$ and put them together, the combined focal length is $c$. What's happening in this case is that the “strength” of the first lens is $1/a$, of the second is $1/b$, and the strengths add.
Another example from optics is that if you have a lens of focal length $c$ and place something a distance of $a$ in front of it, the image forms at a distance of $b$ behind the lens.
A: I don't know what its origin is. I learned that from a book long time ago. And I thought it was used to calculate electrical resistances placed in parallel, which Tpofofn also mentioned.

A: Another non-mathematical example is Sum Frequency Generation in non-linear optics, e.g.

...generation of red light (→ red lasers), e.g. by mixing the outputs
  of a 1064-nm Nd:YAG laser and a 1535-nm fiber laser, resulting in an
  output at 628 nm

http://www.rp-photonics.com/sum_and_difference_frequency_generation.html
Here we have 1/1064 nm + 1/1525 nm = 1/628.41 nm
A: If you have two poles of length $a$ and $b$, and string wire from the top of one pole to the bottom of the other, and vice versa, the height $c$ at which the wires  intersect will be given by half the harmonic mean of the height of the two poles, $\displaystyle \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{b}$, no matter how far apart they are.  This can be easily shown using similar triangles.
I can also think of one more place where the harmonic mean is commonly found in a geometric context:  Given a right triangle with legs $a$, $b$, the squared height of the altitude to the hypotenuse is given by half the harmonic mean of the squared lengths. Wikipedia provides a handy diagram so I'll just quickly write it out:

(Here $A$, $B$, $C$ are the angles opposite the lowercase sides, as is usual, and $P$ is the point where the altitude intersects the hypotenuse.) Then, since $\bigtriangleup ABC \; \sim \; \; \bigtriangleup CBP \;$,
$$\dfrac{a}{f} = \dfrac{c}{b}  \Rightarrow \dfrac{a^2}{f^2} = \dfrac{c^2}{b^2} = \dfrac{a^2 + b^2}{b^2}$$
So, by cross-multiplying we get,
$$\dfrac{1}{f^2} = \dfrac{a^2 + b^2}{a^2b^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}$$
half the harmonic mean of the squared lengths.
A: In a right-angled triangle (with legs a and b, and hypotenuse c) you can inscribe a square in two "obvious" ways:


*

*The square shares the right angle with the triangle and touches c. Then 1/s = 1/a + 1/b

*One side of the square lies on c, and the square touches a and b. Then 1/t = 1/c + 1/h (where h is the height on c)

A: Given three circles of areas $a\geq b\geq c >0$, they can be put in a position such that they are tangent to each other and to a straight line if and only if
$$
\frac{1}{c} = \frac{1}{a} + \frac{1}{b}.
$$

Note that this a special case of Descartes' theorem.
A: Not necessarily geometric, but if $a$ and $b$ are electrical resistances placed in parallel, then their combined resistance is $c$.
