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!
 A: It is also true that $$ \rho \leq 1  $$
In the case that $\rho > 1,$ take $a=1, \; b = -1, \; c = 0.$ The polynomial becomes $2 - 2 \rho = 2(1 - \rho) < 0.$
In a similar style, we can do the original problem this way: if $\rho < -\frac{1}{2},$ take $a=1, \; b = 1, \; c = 1.$ The polynomial becomes $3 + 6 \rho < 3 - 3 = 0.$
Put still another way, the eigenvalues of
$$
\left(
\begin{array}{ccc}
1 & \rho & \rho \\
\rho & 1 & \rho \\
\rho & \rho & 1 \\
\end{array}
\right)
$$
are $$ 1-\rho, \; 1-\rho,  \; 1 + 2 \rho $$
The quadratic form in the question is simply
$$
\left(
\begin{array}{ccc}
a & b & c \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & \rho & \rho \\
\rho & 1 & \rho \\
\rho & \rho & 1 \\
\end{array}
\right)
\left(
\begin{array}{c}
a \\
b \\
c \\
\end{array}
\right)
$$
A: You can show that the inequality is true for all $a,b,c\in\mathbb{R}$ if and only if $-\dfrac{1}{2}\leq \rho \leq 1$.  Indeed, setting $a$, $b$, and $c$ to be $1$ gives
$$3+6\rho\geq 0\,,\text{ or equivalently }\rho\geq -\frac12\,.$$
Taking $(a,b,c)$ to be $(1,0,-1)$ leads to
$$2-2\rho\geq 0\,,\text{ whence }\rho\leq 1\,.$$
We shall prove that the inequality
$$a^2+b^2+c^2+2\rho\,(bc+ca+ab)\geq 0$$
holds for all $\rho\in\left[-\dfrac12,1\right]$.  This is simply because
$$\begin{align}a^2+b^2+c^2+&2\rho\,(bc+ca+ab)\\&=\frac{1+2\rho}{3}\,(a+b+c)^2+\left(\frac{1-\rho}{3}\right)\,\left((b-c)^2+(c-a)^2+(a-b)^2\right)\geq0\,.\end{align}$$
The inequality $a^2+b^2+c^2+2\rho\,(bc+ca+ab)\geq 0$ for $\rho\in\left[-\dfrac12,1\right]$ becomes an equality if and only if


*

*$\rho=-\frac12$ and $a=b=c$,

*$-\frac12<\rho<1$ and $a=b=c=0$, or

*$\rho=1$ and $a+b+c=0$.

A: If you put $a=b=c>0$ we get $$ 1+2\rho \geq 0$$ and thus a conclusion.
A: If $a=b=1$ and $c=x> -1/2$ then we have $$-2\rho\leq {x^2+2\over 2x+1}=:f(x)$$
Since $f$ achieve minimum $1$ at $x=1$ we have $\boxed{\rho\geq-{1\over 2}}$.
And if $x<-{1\over 2}$ we get $$-2\rho\geq {x^2+2\over 2x+1}$$
Since $f$ achieve maximum $-2$ for $x=-3$ we have also $-2\rho \geq -2$ so $\boxed{\rho \leq 1}$. 
A: With $M = \left(
\begin{array}{ccc}
 0 & \frac{1}{2} & \frac{1}{2} \\
 \frac{1}{2} & 0 & \frac{1}{2} \\
 \frac{1}{2} & \frac{1}{2} & 0 \\
\end{array}
\right)$ and $v = (a,b,c)$ we have
$$
v^{\top}I_3 v + 2\rho v^{\top}M v \ge 0\Rightarrow v^{\top}\left(I_3+2\rho M\right)v \ge0
$$
so choosing $\rho$ such that $I_3+2\rho M \gt 0$ (definite positiveness)
with the conditions in the sub-determinants
$$
1-\rho^2 > 0\\
(\rho -1)^2 (2 \rho +1) > 0
$$
follows in $$\rho \ge -\frac 12$$
A: 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 $.
