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 ${{x}_{1}}{{x}_{2}}<0$). 
 A: Observe that 
$$
\left(x_1+\alpha x_2\right)^2 = x_1^2+2\alpha x_1 x_2 +\alpha^2x_2^2=x_1^2+2\alpha x_1 x_2 +\left(\alpha^2-1\right)x_2^2 +x_2^2
$$
and so 
$$
x_1^2+x_2^2+2\alpha {{x}_{1}}{{x}_{2}}=\left(x_1+\alpha x_2\right)^2 -\left(\alpha^2-1\right)x_2^2
$$
substituting into original expression $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+2\alpha {{x}_{1}}{{x}_{2}}>0$ we get
$$
0<x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+2\alpha {{x}_{1}}{{x}_{2}} = 
 \left(x_1+\alpha x_2\right)^2 +x_3^2 -\left(\alpha^2-1\right)x_2^2
$$
The last expression (inequality) can be rewritten in the form
$$
\left(x_1+\alpha x_2\right)^2 +x_3^2 >\left(\alpha^2-1\right)x_2^2
$$
Since both terms in the left hand side are non-negative, the inequality will hold true for any values of $x_1,x_2,$ and $x_3$ if and only if right hand side is negative, i.e. 
$$\alpha^2-1<0\iff\alpha^2<1\iff\left\lvert\alpha\right\rvert<1.$$
A: 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 $|\alpha|<1$.
A: $2\alpha x_1x_2>-(x_1^2+x_2^2+x_3^2)$
RHS must be smaller than $0$, if LHS>$0$, the equation must work. We only consider LHS<$0$
Then we have,
$-2|\alpha| |x_1||x_2|>-(x_1^2+x_2^2+x_3^2)$
$\Leftarrow2|\alpha| |x_1||x_2|<|x_1|^2+|x_2|^2+|x_3|^2$
$\Leftarrow|\alpha|<\frac{|x_1|^2+|x_2|^2+|x_3|^2}{2 |x_1||x_2|}$
$\frac{|x_1|^2+|x_2|^2+|x_3|^2}{2 |x_1||x_2|}\geqslant1 $ $\Leftarrow$ $|x_1|^2+|x_2|^2+|x_3|^2\geqslant|x_1|^2+|x_2|^2\geqslant2 |x_1||x_2|$$\Leftarrow(|x_1|-|x_2|)^2\geqslant0$
If $|\alpha|>1,$ when $|x_3|=0,|x_1|=|x_2|$,$\frac{|x_1|^2+|x_2|^2+|x_3|^2}{2 |x_1||x_2|}=1<|\alpha| $. Contradict, so $|\alpha|<1$.
Inversely, if $|\alpha|<1,$$\frac{|x_1|^2+|x_2|^2+|x_3|^2}{2 |x_1||x_2|}\geqslant1 > |\alpha|$ 
Q.E.D
