<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 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 $1+1-1=1$.
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