$a^2 + b^2 + c^2 = 1 ,$ then $ab + bc + ca$ gives =? In a recent examination this question has been asked, which says:
$a^2+b^2+c^2 = 1$ , then  $ab + bc + ca$ gives = ?
What should be the answer? I have tried the formula for $(a+b+c)^2$, but gets varying answer like $0$ or $0.25$, on assigning different values to variables.
How to approach such question? 
 A: Just knowing that $a^2+b^2+c^2=1$ is not enough to determine the value of $ab+bc+ca$. For example, if $a=b=0$ and $c=1$, then $ab+bc+ca = 0$. On the other hand, if $a=b=c=\frac{1}{\sqrt3}$, then $ab+bc+ca = 1$. In fact, using $$(a+b+c)^2=a^2+b^2+c^2 + 2(ab+bc+ca),$$ you get that $$ab+bc+ca=\frac{(a+b+c)^2 - 1}{2}.$$
A: It depends on $a+b+c$:
$$
(a+b+c)^2-2(ab+bc+ca)=1\\
ab+bc+ca=\dfrac{(a+b+c)^2-1}{2}
$$
A: $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1+2(ab+bc+ca)$
therefore $$(ab+bc+ca)=((a+b+c)^2-1)/2$$ so the value is not fixed
A: With this information the best you can do is the following
$$ab+bc+ac\leq1$$
which is a direct result of the AM-GM inequality i.e.
$$a^2+b^2\geq2ab$$
$$b^2+c^2\geq2bc$$
$$a^2+c^2\geq2ac$$
Summing sides respectively
$$1=a^2+b^2+c^2\geq ab+bc+ac$$
A: $$ 
(a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab + bc + ca)
$$
The LHS of the above identity is a perfect square, hence it is always positive or 0.
Thus,
$$
a^2 + b^2 +c^2 + 2(ab + bc + ca) ≥ 0
$$
It is given that $a^2 + b^2 +c^2 =1 $ .
$$
1 + 2(ab + bc + ca) ≥ 0
$$
Therefore,
$$
ab + bc + ca ≥ -1/2
$$
It is possible to determine the upper bound of the above expression by applying the AM-GM inequality as follows: (as already done by Arian)
$$
a^2+b^2≥2ab\\
b^2+c^2≥2bc\\
a^2+c^2≥2ac
$$
Adding the above three inequalities, we get
$$
a^2 + b^2 +c^2  ≥ ab + bc + ca
$$
which implies
$$
ab + bc + ca ≤ 1
$$
Thus, the value of $ab + bc + ca$ lies in the interval $[-1/2, 1]$ .
Note: It is not possible to find an exact value for this expression, as already noted by the other answers.
A: I believe your answers should be in terms of a,b,c
(a+b+c-1)(a+b+c+1)/2
