If a² + b² + c² = 1, then ab + bc + ca lies in the interval (A) [1/2, 2] (B) [-1, 2] (C) [-1/2, 1] (

Q: Wait, why can't we just pick values like $a=1, b=0, c=0$ to test the range?

That is a classic trap! Testing specific points like $(1, 0, 0)$ gives you $ab+bc+ca = 0$, which is inside the interval, but it doesn't tell you the *boundaries*. To find the full range, you must use algebraic inequalities that hold for *all* possible values of $a, b, c$ that satisfy the condition. Testing points only gives you a subset of the range, not the complete interval.

Q: How did you know to use $(a - b)^2 + (b - c)^2 + (c - a)^2 \ge 0$ for the upper bound?

Great question! In JEE problems, when you see symmetric expressions involving squares, the sum of squares of differences is a powerful tool. Since $(a - b)^2$, $(b - c)^2$, and $(c - a)^2$ are all non-negative, their sum must be $\ge 0$. This identity is a standard 'trick' to relate the sum of squares to the sum of cross-products. It's a fundamental part of your algebraic toolkit.

Q: Is it possible for the expression to be less than $-1/2$?

No, it is not. Our derivation $(a + b + c)^2 \ge 0$ proves that $1 + 2(ab + bc + ca) \ge 0$, which simplifies strictly to $ab + bc + ca \ge -1/2$. Because this inequality is derived from a universal property of real numbers, it is impossible for the expression to fall below $-1/2$ while satisfying the given condition.

Question

If $a^2 + b^2 + c^2 = 1$, then $ab + bc + ca$ lies in the interval

Options:

(A) $[1/2, 2]$

(B) $[-1, 2]$

(C) $[-1/2, 1]$

(D) $[-1, 1/2]$

Sigma Percentile · Accepted Answer

### The Beauty of Algebraic Constraints

Welcome, fellow traveler on the JEE journey! Today, we are going to peel back the layers of a problem that, at first glance, seems like a simple algebraic exercise but is actually a beautiful dance of inequalities.

We are given the condition $a^2 + b^2 + c^2 = 1$ and asked to find the range of $ab + bc + ca$. This is not just about finding an answer; it is about understanding how constraints shape the behavior of variables.

### The Lower Bound: The Power of the Perfect Square

To find the lower bound, we need to connect our target expression to the given condition. Think about the identity for the square of a sum:

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

We know that for any real numbers $a, b, c$, the square of their sum must be non-negative, so $(a + b + c)^2 \ge 0$. This is our starting point.

By substituting our given condition $a^2 + b^2 + c^2 = 1$ into this identity, we get:

$$ 1 + 2(ab + bc + ca) \ge 0 $$

Now, it is just a matter of simple algebra. Subtracting $1$ from both sides gives $2(ab + bc + ca) \ge -1$, and dividing by $2$ reveals our lower bound:

$$ ab + bc + ca \ge -\frac{1}{2} $$

This is the absolute floor for our expression.

### The Upper Bound: The Hidden Symmetry

Now, how do we find the upper bound? We need another universal truth. Consider the sum of the squares of the differences: $(a - b)^2 + (b - c)^2 + (c - a)^2$.

Since each term is a square, the entire sum must be greater than or equal to zero. Let's expand this:

$$ (a^2 - 2ab + b^2) + (b^2 - 2bc + c^2) + (c^2 - 2ca + a^2) \ge 0 $$

If we group the terms, we see that $a^2$, $b^2$, and $c^2$ each appear twice. This simplifies to:

$$ 2(a^2 + b^2 + c^2) - 2(ab + bc + ca) \ge 0 $$

Again, we substitute our given condition $a^2 + b^2 + c^2 = 1$. This transforms the inequality into:

$$ 2(1) - 2(ab + bc + ca) \ge 0 $$

Rearranging this, we get $2 \ge 2(ab + bc + ca)$, which simplifies to $1 \ge ab + bc + ca$. There is our upper bound!

### The Final Synthesis

We have successfully trapped our expression between two walls. We know it cannot be less than $-\frac{1}{2}$ and it cannot be greater than $1$.

Therefore, the expression $ab + bc + ca$ must lie in the closed interval:

$$ \left[ -\frac{1}{2}, 1 \right] $$

This problem is a perfect example of how, in the world of JEE Advanced, we don't just calculate; we use the inherent properties of numbers to define the boundaries of possibility. Keep practicing these identities, and soon, you will see them everywhere!

Animated Solution for Mathematics - Quadratic Equations: If $a^{2} + b^{2} + c^{2} = 1$ , then $ab + b c + c a$ lies in the interval

Select Answer:

Visualized Solution

Given Condition and Objective

The Lower Bound Logic

Expanding the Identity

Substituting the Given Value

Solving for the Lower Bound

The Upper Bound Logic

Expanding the Differences

Substituting Again

Solving for the Upper Bound

Final Conclusion

The Sigma Insight: Maximum and Minimum Values of Quadratic Expressions

The Beauty of Algebraic Constraints

The Lower Bound

The Upper Bound

The Final Synthesis

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a2+b2+c2=1, then ab+bc+ca lies in the interval