Step 1: State what we need to determine.
Given $a, b, c$ are distinct positive reals with $a^2+b^2+c^2=1$, we want to find the relationship between $ab+bc+ca$ and $1$. The key idea is to use a well-known algebraic inequality that connects the sum of squares with the sum of products.
Step 2: Start from the always-non-negative sum of squared differences.
For any real numbers, $(a-b)^2 \geq 0$, $(b-c)^2 \geq 0$, $(c-a)^2 \geq 0$. Adding all three: $(a-b)^2+(b-c)^2+(c-a)^2 \geq 0$.
Step 3: Expand the left side and rearrange.
$(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2) \geq 0$ gives $2(a^2+b^2+c^2)-2(ab+bc+ca) \geq 0$, so $a^2+b^2+c^2 \geq ab+bc+ca$.
Step 4: Strengthen to strict inequality using distinctness.
Since $a, b, c$ are all distinct, no two are equal. So $a \neq b$ implies $(a-b)^2 > 0$. Therefore at least one squared difference is strictly positive, making the total sum strictly positive: $(a-b)^2+(b-c)^2+(c-a)^2 > 0$. This forces $a^2+b^2+c^2 > ab+bc+ca$.
Step 5: Substitute the given constraint.
Replacing $a^2+b^2+c^2 = 1$: $1 > ab+bc+ca$.
Step 6: State the final answer.
The value of $ab+bc+ca$ is less than $1$. \[ \boxed{ab+bc+ca < 1} \]