How find this value $\frac{a^2+b^2-c^2}{2ab}+\frac{a^2+c^2-b^2}{2ac}+\frac{b^2+c^2-a^2}{2bc}$ let $a,b,c$ such that
$$\left(\dfrac{a^2+b^2-c^2}{2ab}\right)^2+\left(\dfrac{b^2+c^2-a^2}{2bc}\right)^2+\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2=3,$$
find the value 
$$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}$$
is true?
Yes, I tink this problem can prove
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$
so $$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}=1or -3$$
How many nice methods prove $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$ ?
and  I  see this easy problem
http://zhidao.baidu.com/question/260913315.html
 A: First, we simplify the initial equation by multiplying out by the denominator. Let
$$f(a,b,c) = c^2(a^2+b^2-c^2)^2 + a^2(b^2+c^2-a^2)^2 + b^2(c^2+a^2-b^2)^2 - 12a^2b^2c^2$$
In Ron Gordon's deleted post, he realized that $f(a,b,c) = 0 $ if $a+b=c$, $b+c=a$, $c+a=b$. This strongly suggests that $a+b-c, b+c-a, c+a-b$ are factors (sort of Remainder Factor Theorem). And indeed they are. We have
$$f(a,b,c) = -(a+b-c)(a-b+c)(-a+b+c)(a+b+c)(a^2+b^2+c^2)$$
which you can check in Wolfram.
Since the denominators are non-zero, we have $(a^2+b^2+c^2)>0$. Thus $$f(a,b,c) = 0 \Leftrightarrow (a+b-c)(a-b+c)(-a+b+c)(a+b+c) = 0 $$
We now split into cases.
Case 1. $(a+b-c)(a-b+c)(-a+b+c) = 0$
(Once again, multiply by denominators, and using Ron's observation.) Defining
$$g(a,b,c) = c(a^2+b^2-c^2) + a(b^2+c^2-a^2) + b(c^2+a^2-b^2) - 2abc $$
gives us 
$$g(a,b,c) = - (a+b-c)(a-b+c)(-a+b+c) = 0 $$
Hence, the answer is 1.
Case 2. $a+b+c = 0$.
Then each term can be simplified in the form $\frac{ (a+b)^2 - c^2 - 2ab}{2ab} = -1$, hence the answer is -3.
A: I think the answer must be $1$ if we impose a triangle inequality.  The individual terms are all cosines of angles of a triangle.  Thus, the sum of those angles must be $\pi$.  But the sum of the squares of their cosines is $3$; therefore each cosine must be $\pm 1$.  But the sum of the angles is, again, $\pi$, so that two of the angles must be zero and the third $\pi$.  Thus, the sum of the cosines, which is sought, must be $1+1-1=1$.
A: I think you might be talking about a triangle, with $a$, $b$, $c$ as it's sides of a triangle. If the angles opposite to side a, b, c are A, B,C, then $$\cos A=\frac{b^2+c^2-a^2}{2cb}$$ and similarly, $$\cos B=\frac{a^2+c^2-b^2}{2ac}$$ and $$\cos C=\frac{b^2+a^2-c^2}{2ab}$$ If it is true then $\cos A = \cos B = \cos C =1$ from the first term (since minimum and maximum of cos x with any x is plus minus 1). So the answer is either -3, -2,-1 ,1,2 or 3. 
Form here, we are getting a variety of answers, but since here only one of the angle can be negative(obtuse), the answer could only be 1 or 3. (since the sum could only be 1+1-1 or 1+1+1).
Now since all three angles of the triangle cannot be 0, so they must be one angle 180 and the rest 0. So 1 is the only answer.
NOTE: The above is only true when a,b and c are angles of a triangle. And the above is the called the cosines rule of a triangle. Although the triangle discussed above is in fact a straight line, but it can be noticed that the sum of the two sees containing the 180 angle is equal to the third side.
