Why is $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+2\alpha {{x}_{1}}{{x}_{2}}>0$ for all $({{x}_{1}},{{x}_{2}},{{x}_{3}})\ne (0,0,0)$ if and only if $|\alpha |<1$? Why is $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+2\alpha {{x}_{1}}{{x}_{2}}>0$ for all $({{x}_{1}},{{x}_{2}},{{x}_{3}})\ne (0,0,0)$  if and only if $|\alpha |<1$? I tried solving the inequality and just ended up with  $\alpha >-\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}{2{{x}_{1}}{{x}_{2}}}$ (if ${{x}_{1}}{{x}_{2}}>0$ ) and $\alpha <-\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}{2{{x}_{1}}{{x}_{2}}}$(if ${{x}_{1}}{{x}_{2}}<0$). 

There are probably easier ways to do it. But whatever, consider the matrix representing this quadratic form: $$A = \begin{pmatrix} 1 & \alpha & 0 \\ \alpha & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}.$$Your condition is equivalent to the matrix $A$ being positive-definite, which by Sylvester's Criterion happens if and only if all principal minors are positive. We compute $$\begin{align} \Delta_1 &= 1, \\ \Delta_2 &= 1-\alpha^2, \\ \Delta_3 &= 1-\alpha^2.  \end{align}$$These guys are all positive if and only if $1-\alpha^2 > 0$, which is the same as $|\alpha|<1$.
