Prove or disprove $|x+y+z|\ge\sqrt{3}$ Let $x,y,z\in R$, such $|x|\neq|y|\neq|z|$,and $|x|,|y|,|z|&gt;1$ and $xy+yz+xz=-1$
prove or disprove
$$|x+y+z|\ge\sqrt{3}$$

I try

$$(x-y)^2+(y-z)^2+(z-x)^2\ge 0$$
so $$|x+y+z|^2\ge 3(xy+yz+xz)=-3$$
why?
Counterexample:

Suppose $x = 1.02, y = -1.0001, z = \frac {-1-xy}{x+y} = \frac {0.020102}{0.0199} \approx 1.01015$

$|x|,|y|,|z| &gt; 1$
and
$xy + yz + xz = -1$
and $|x+y+z| = 1.03 &lt; \sqrt 3$

Since $|x|, |y|, |z| &gt; 1, x^2 + y^2 + z^2 &gt; 3$

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) \ge 0\\
(x+y+z)^2 \ge 3 -2 \ge 0\\
(x+y+z)^2 \ge 1\\
|x+y+z| \ge 1$
Counterexample:

Suppose $x = 1.02, y = -1.0001, z = \frac {-1-xy}{x+y} = \frac {0.020102}{0.0199} \approx 1.01015$

$|x|,|y|,|z| &gt; 1$
and
$xy + yz + xz = -1$
and $|x+y+z| = 1.03 &lt; \sqrt 3$

Since $|x|, |y|, |z| &gt; 1, x^2 + y^2 + z^2 &gt; 3$

$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) \ge 0\\
(x+y+z)^2 \ge 3 -2 \ge 0\\
(x+y+z)^2 \ge 1\\
|x+y+z| \ge 1$