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!

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
$$
\left(
\begin{array}{ccc}
a & b & c \\
\end{array}
\right)
\left(
\begin{array}{ccc}
1 & \rho & \rho \\
\rho & 1 & \rho \\
\rho & \rho & 1 \\
\end{array}
\right)
\left(
\begin{array}{c}
a \\
b \\
c \\
\end{array}
\right)
$$
