Quadratic Equations Question - None

Question

If $a, b, c$ are distinct positive real numbers and $a^2 + b^2 + c^2 = 1$ then $ab + bc + ca$ is

Accepted Answer

Step 1: Consider $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$

Step 2: Since $a^2 + b^2 + c^2 = 1$:

$(a + b + c)^2 = 1 + 2(ab + bc + ca)$

Step 3: Since $a, b, c > 0$, we have $(a + b + c)^2 > 0$

Step 4: Consider $(a - b)^2 + (b - c)^2 + (c - a)^2 > 0$ (since they are distinct)

Expanding: $2(a^2 + b^2 + c^2) - 2(ab + bc + ca) > 0$

Step 5: $2(1) - 2(ab + bc + ca) > 0$

$ab + bc + ca < 1$

∴ Answer is (A).

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