Let $x,y,z\in R$, such $|x|\neq|y|\neq|z|$,and $|x|,|y|,|z|>1$ 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?
Prove or disprove $|x+y+z|\ge\sqrt{3}$
2
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inequality
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1Take $x=1+\epsilon$, $y=1+2\epsilon$ and $z = \frac{-1-xy}{x+y} = -1 - \epsilon^2 + O(\epsilon^3)$ then $xy+yz+zx = -1$ and $|x+y+z| = 1 + 3\epsilon + O(\epsilon^2)$ so any value larger than $1$ is possible. As in the now deleted answer you can prove that $|x+y+z| > 1$ always holds. – 2017-01-12
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1The inequality fails for $x=-y=z=1$ so by continuity it will also fail in a small enough neighborhood thereof where $|x|,|y|,|z|$ are strictly $>1$. – 2017-01-12
1 Answers
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Counterexample:
Suppose $x = 1.02, y = -1.0001, z = \frac {-1-xy}{x+y} = \frac {0.020102}{0.0199} \approx 1.01015$
$|x|,|y|,|z| > 1$ and $xy + yz + xz = -1$ and $|x+y+z| = 1.03 < \sqrt 3$
Since $|x|, |y|, |z| > 1, x^2 + y^2 + z^2 > 3$
$(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) \ge 0\\ (x+y+z)^2 \ge 3 -2 \ge 0\\ (x+y+z)^2 \ge 1\\ |x+y+z| \ge 1$
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0I didn't downvote and I am just commenting to understand something from your answer that isn't clear to me. Can you explain how the third equation (from the last) implies the second (again from the last)? – 2017-01-12
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0You are right, I really chunked it up... – 2017-01-12
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0@user170039 this is better – 2017-01-12