Question

If \(a\), \(b\), \(c\) are distinct positive real numbers and \[ a^2+b^2+c^2=1, \] then the value of \(ab+bc+ca\) is

• A. less than \(1\)
• B. greater than \(1\)
• C. equals to \(1\)
• D. any real number

Hint:

Use the identity \[ (a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2-ab-bc-ca) \] to compare \[ a^2+b^2+c^2 \] and \[ ab+bc+ca. \]

Answer 1

The Correct Option is A

Step 1: Use the given condition.
Given, \[ a^2+b^2+c^2=1 \] We need to determine the nature of \[ ab+bc+ca \]

Step 2: Use a standard identity.
We know that \[ (a-b)^2+(b-c)^2+(c-a)^2\geq 0 \] Expanding, \[ a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\geq 0 \] \[ 2(a^2+b^2+c^2)-2(ab+bc+ca)\geq 0 \] Dividing by \(2\), \[ a^2+b^2+c^2\geq ab+bc+ca \]

Step 3: Apply distinctness of \(a\), \(b\), \(c\).
Since \(a\), \(b\), \(c\) are distinct, at least one of \[ (a-b)^2,\quad (b-c)^2,\quad (c-a)^2 \] is positive.
Therefore, \[ (a-b)^2+(b-c)^2+(c-a)^2\gt 0 \] So, \[ a^2+b^2+c^2\gt ab+bc+ca \]

Step 4: Substitute the given value.
Since \[ a^2+b^2+c^2=1, \] we get \[ 1\gt ab+bc+ca \] Thus, \[ ab+bc+ca\lt 1 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\text{less than }1} \]