if $a^2+b^2+c^2+2\rho(ab+bc+ca)\ge0$ then $\rho\ge-1/2$

Question

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!

Answer 1

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

Answer 2

If you put $a=b=c>0$ we get $$ 1+2\rho \geq 0$$ and thus a conclusion.

Answer 3

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

Answer 4

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 $a+b+c=0$.

Answer 5

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 $$\rho \ge -\frac 12$$

Answer 6

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 $.