<s>[INST] 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
 [/INST] 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.
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