Prove the identity:
(hint: use the laws of cosine)

cos A
= b² + c² - a²
..a.........2abc

Using the above fact, now prove:
(Hint: what would cos B and cos C look like?)
....................b.........c

cos A + cos B + cos C = a² + b² + c²
..a.......b.......c........2abc

Prove the identity:
(hint: use the laws of cosine)

cos A = b² + c² - a²
..a.........2abc

Using the above fact, now prove:
(Hint: what would cos B and cos C look like?)
....................b.........c

cos A + cos B + cos C = a² + b² + c²
..a.......b.......c........2abc

The Law of cosines is $\displaystyle a^2 = b^2+c^2-2bc\cos A$

Re-arranging: $\displaystyle -2bc\cos A = a^2-b^2-c^2$

Multiplying both sides by -1: $\displaystyle 2bc\cos{A}=b^2+c^2-a^2$

Dividing both sides by 2abc: $\displaystyle \frac{\cos{A}}{a}=\frac{b^2+c^2-a^2}{2abc}$

Similarly, by changing around variables, you can get:

$\displaystyle \frac{\cos B}{b}=\frac{a^2+c^2-b^2}{2abc}$

$\displaystyle \frac{\cos C}{c} = \frac{a^2+b^2-c^2}{2abc}$

So if you simply add them all up:

$\displaystyle \frac{\cos A}{a}+\frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2+c^2-a^2}{2abc}+\frac{a^2+c^2-b^2}{2abc}+\frac{a^2+b^2-c^2}{2abc}$

$\displaystyle =\frac{a^2+b^2+c^2}{2abc}$

3. Fantastic, thank you very much!