<s>[INST] $a^2 + b^2 + c^2 = 1 ,$ 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? 
 [/INST] $$ 
(a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab + bc + ca)
$$
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.
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