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

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?

• The question is fine, and the method is fine, too, but you should be a little more careful: start with $(a+b+c)^2\ge 0$. – W-t-P Apr 26 at 9:06
• When $a=\frac{1}{\sqrt 2},b=-\frac{1}{\sqrt 2},c=0$, we get $ab+bc+ca\color{red}=-\frac 12$. – mathlove Apr 26 at 9:08
• Yes equality sign is Missing in the answer, I will correct it. – sawan kumawat Apr 26 at 9:15
• $$\displaystyle -\frac{1}{2}\leq (ab+bc+ca)\leq 1$$ – DXT Apr 26 at 9:20

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 $$a+b+c=0,\,a^2+b^2+c^2=1$$.
• I got it, I was missing $(a+b+c)^2 \ge 0$ in every real number case – sawan kumawat Apr 26 at 9:25