<s>[INST] if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ then $\rho\ge-1/2$ This may be trivial but I am not able to prove that if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ for $a,b,c\in\mathbb{R}$ then $\rho\ge-1/2$. Can anybody help me please?
Thanks!
 [/INST] Correct me if wrong:
Consider $(a,b,c)$ , then
1) $a^2+b^2+c^2=$
$ (a,b,c)\cdot(a,b,c)=||(a,b,c)||^2$.
2) $ab+bc +ca =$
$(a,b,c) \cdot (b,c,a) =$
$||(a,b,c)|| ||(a,b,c)|| \cos \phi =$
$||(a,b,c)||^2 \cos \phi$.
Combining:
$a^2+b^2+c^2 +2\rho (ab+bc+cd) \ge$
$0$, or
$||(a,b,c)||^2 + 2\rho ||(a,b,c)||^2 \cos \phi \ge 0.$
Assuming $(a,b,c)\not = (0,0,0)$:
$1+ 2\rho \cos \phi \ge 0$;
$\rho \cos \phi \ge -1/2$.
1) $0 \le \cos \phi \le 1$, then
$\rho \ge -1/2$. 
2) $-1/2 \le \cos \phi \le 0$, then
$\rho \le 1$.
Note : $\cos \phi \ge -1/2$, since
$(a+b+c)^2 - (a^2+b^2+c^2) = $
$2(ab +ac+bc)$.
We have:
$-||(a,b,c)||^2 \le $
$(a+b+c)^2 -||(a,b,c)||^2 =$
$ 2||(a,b,c)||^2\cos \phi.$
Hence: $-1 \le 2\cos \phi $.
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