<s>[INST] Law of Cosines and Heron's Formula in inequalities I had the following question:

Suppose $a$, $b$, and $c$ are non-zero real numbers, and $x$, $y$, and $z$ satisfy the equations
  $$
bx + ay = c, cx + az = b, cy + bz = a.
$$
  Prove that $-1 < x, y, z < 1$ if and only if $a^4 + b^4 + c^4 < 2 (a^2b^2 + b^2c^2 + c^2a^2)$.

After some thoughts, I factorized the inequality to $(a+b+c)(-a+b+c)(a-b+c)(a+b-c)>0$, which I believe is related to Heron's Formula of area.
I also rewrote the three equations into $a^2=b^2+c^2-2bcx$, $b^2=a^2+c^2-2acy$ and $c^2=a^2+b^2-2abz$, which I believe is related to the Law of Cosine.
I am guessing that $a$, $b$ and $c$ are sides of triangles and $x$, $y$, $z$ are the cosine value to the angle opposite the sides. However, I wasn't able to do this without assuming the side lengths are positive. 
 [/INST] I hope the following can help.
Since $$0<a^2b^2c^2=ab\cdot ac\cdot bc,$$ we can assume that $$bc>0.$$
Thus, since $-1<x<1$ and by your work $x=\frac{b^2+c^2-a^2}{2bc},$ we obtain:
$$-1<\frac{b^2+c^2-a^2}{2bc}<1,$$ which gives
$$(a+b+c)(b+c-a)>0$$ and $$(a+b-c)(a+c-b)>0$$ and from here
$$(a+b+c)\prod_{cyc}(a+b-c)>0$$ or
$$\sum_{cyc}(2a^2b^2-a^4)>0.$$
Also, let $\sum\limits_{cyc}(2a^2b^2-a^4)>0,$ $abc\neq0$, $x=\frac{b^2+c^2-a^2}{2bc},$ $y=\frac{a^2+c^2-b^2}{2ac}$ and $z=\frac{a^2+b^2-c^2}{2ab}.$
We need to prove that $$\{x,y,z\}\subset(-1,1).$$
Indeed, changing $a$ on $-a$ does not change the condition $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)>0$$ and what we need to prove.
Thus, we can assume that $a>0$, $b>0$ and $c>0$ and the rest is smooth.
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