Prove that $(a b+b c+c a-1)^{2} \leq\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)$. 
Let $a$, $b$, and $c$ be real numbers.  Prove that $$(a b+b c+c a-1)^{2} \leq\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)\,.$$

In solution of this author take Let $a=\tan x, b=\tan y, c=\tan z$ with $-\frac{\pi}{2}<x, y, z<\frac{\pi}{2}$ but i did not understand the reason behind letting that $-\frac{\pi}{2}<x, y, z<\frac{\pi}{2}$ ,
i mean if we just take  $a=\tan x, b=\tan y, c=\tan z$ then is something wrong,
I do not want solution,just want to clear this step

Since $$(x^2+y^2)(z^2+t^2)=(xz+yt)^2+(xt-yz)^2,$$ we obtain:
$$(a^2+1)(b^2+1)(c^2+1)=((a+b)^2+(ab-1)^2)(c^2+1)=$$
$$=((a+b)c+ab-1)^2+(a+b-(ab-1)c)^2\geq(ab+ac+bc-1)^2.$$
