<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] 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 $a+b+c=0$.

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