# If <mtable columnalign="right center left" rowspacing="3pt" columnspacing="0 thickmathspace" di

If
$\begin{array}{r}x=cy+bz\\ y=az+cx\\ z=bx+ay\end{array}$
where $x,y,z$ are not all zero, prove that ${a}^{2}+{b}^{2}+{c}^{2}+2abc=1$
Further if at least one of $a,b,c$ is a proper fraction, prove that:
(i) ${a}^{2}+{b}^{2}+{c}^{2}<3$
(ii) $abc>-1$
You can still ask an expert for help

## Want to know more about Inequalities systems and graphs?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Kaydence Washington
Assuming $a$ is a proper fraction, then
${a}^{2}<1$
Considering quadratic in $c$,
${c}^{2}+2abc+{a}^{2}+{b}^{2}-1=0$
Admitting real value of $c$,$\begin{array}{rl}\mathrm{\Delta }& \ge 0\\ 4\left({a}^{2}-1\right)\left({b}^{2}-1\right)& \ge 0\\ {b}^{2}-1& \le 0\end{array}$
Hence,
${b}^{2}\le 1$
Similarly,
${c}^{2}\le 1$
Hence,
${a}^{2}+{b}^{2}+{c}^{2}<3$
Also
$abc=\frac{1-{a}^{2}-{b}^{2}-{c}^{2}}{2}>-1$