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If $ a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = 1$, then show that $ab+bc+ca\ge \frac{-1}{2}$

If figured out that if I put $(a+b+c)^2 = 0$ then I will get the above answer, but $(a+b+c)^2 = 0$ is not given in the question, so is there any other method to do it, or the question is wrong?

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    $\begingroup$ The question is fine, and the method is fine, too, but you should be a little more careful: start with $(a+b+c)^2\ge 0$. $\endgroup$ – W-t-P Apr 26 at 9:06
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    $\begingroup$ When $a=\frac{1}{\sqrt 2},b=-\frac{1}{\sqrt 2},c=0$, we get $ab+bc+ca\color{red}=-\frac 12$. $\endgroup$ – mathlove Apr 26 at 9:08
  • $\begingroup$ Yes equality sign is Missing in the answer, I will correct it. $\endgroup$ – sawan kumawat Apr 26 at 9:15
  • $\begingroup$ $$\displaystyle -\frac{1}{2}\leq (ab+bc+ca)\leq 1$$ $\endgroup$ – DXT Apr 26 at 9:20
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You should use $0\le(a+b+c)^2=1+2(ab+bc+ca)$. Your inequality actually shouldn't be strict, because e.g. we can take $a=0,\,b=-c=\frac{1}{\sqrt{2}}$ to get $a+b+c=0,\,a^2+b^2+c^2=1$.

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  • $\begingroup$ I got it, I was missing $(a+b+c)^2 \ge 0$ in every real number case $\endgroup$ – sawan kumawat Apr 26 at 9:25

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