Prove that $(a b+b c+c a-1)^{2} \leq\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)$.

Question

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

Answer 1

The range of $\tan x$ over $x\in(-\frac{\pi}2,\frac{\pi}2)$ is $\mathbb R$, so we need to take $x$ as is said because $a\in\mathbb R$ is given. Take the domain any less, and there exists $a\in\mathbb R$ such that $a\neq\tan x$ for any $x$ in the domain. Same goes for $b,c$.

Answer 2

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

Answer 3

We have $$(a^2+1)(b^2+1)(c^2+1)-(ab+bc+ca-1)^2=(a+b+c-abc)^2\ge 0$$