Minimum value of $ab+bc+ca$ depending on given constraints We have to minimize the function $ab+bc+ca$ on the given condition that $a^2+b^2+c^2=1$.
I have tried the following approach (but I don't think I'm right )
$$\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)
$$
We know that all squares are greater than equal to 0 and hence:  $\left(a+b+c\right)^2\geq0$
$$\therefore  a^2+b^2+c^2+2\left(ab+bc+ca\right) \geq0$$
$$\Rightarrow 1+2\left(ab+bc+ca\right) \geq0$$
$$\Rightarrow 2\left(ab+bc+ca\right)\geq -1$$
$$\Rightarrow \left(ab+bc+ca\right)\geq -1/2$$
As we can see I have reached a definite answer but my question is since there is a given constraint on $a$,$b$ and $c$ we can't say for sure that this value will be greater than or equal to -1/2, or can we?

You're doing it right. Another way of looking at what you're doing:
You have
$$ab + bc + ac = {1 \over 2} (a + b + c)^2 - {1 \over 2}(a^2 + b^2 + c^2)$$
Under the constraint that $a^2 + b^2 + c^2$, you therefore have
$$ab + bc + ac = {1 \over 2} (a + b + c)^2 - {1 \over 2}$$
You're trying to minimize this quantity. Since ${1 \over 2} (a + b + c)^2 \geq 0$, it's smallest when $a + b + c = 0$, in which case you have
$$ab + bc + ac = -{1 \over 2}$$
This is the minimum possible value of $ab + bc + ac$, and it is achieved for any $(a,b,c)$ for which $a^2 + b^2 + c^2 = 1$ and $a + b + c = 0$. There are many such $(a,b,c)$ as they correspond to the intersection of the sphere $a^2 + b^2 + c^2 = 1$ and the plane through the origin with equation $a + b + c = 0$.
