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!

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 $\boxed{\rho \leq 1}$. 
