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Animated Solution for Mathematics - Basic Mathematics: If $a, b, c$ are distinct +ve real numbers and $a^{2} + b^{2} + c^{2} = 1$ then $ab + b c + c a$ is

Q: Wait, why do we reject the 'equal to' case? Why is it strictly less than 1?

That is a brilliant observation! The problem explicitly states that $a, b,$ and $c$ are distinct. If $a = b = c$, then $(a-b)^2$ would be zero. But because they are distinct, $(a-b)^2$ must be strictly positive. This forces the entire sum to be strictly greater than zero, which propagates through the algebra to make our final result strictly less than 1.

Q: Can't we just use the AM-GM inequality here?

You absolutely can! By AM-GM, we know that $a^2 + b^2 \ge 2ab$, $b^2 + c^2 \ge 2bc$, and $c^2 + a^2 \ge 2ca$. Adding these gives $2(a^2 + b^2 + c^2) \ge 2(ab + bc + ca)$, which simplifies to $ab + bc + ca \le a^2 + b^2 + c^2 = 1$. However, AM-GM equality holds only when $a=b=c$. Since the problem specifies they are distinct, we cannot reach the equality case, hence the strict inequality.

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry of the problem. When you see

a^{2} + b^{2} + c^{2}

and

ab + b c + c a

, your mind should immediately jump to the expansion of

(a - b)^{2} + (b - c)^{2} + (c - a)^{2}

. Why? Because this expression is always non-negative, and it perfectly bridges the gap between the sum of squares and the sum of pairwise products.

Hint 2

Hint 2: Remember the constraint that

a, b,

and

c

are distinct. This is not just a piece of information; it is the key to the lock. If they were equal, the sum of squared differences would be zero. Since they are distinct, that sum must be strictly greater than zero. This is where the strict inequality comes from.

Hint 3

Hint 3: Start your algebraic journey by writing down the inequality:

(a - b)^{2} + (b - c)^{2} + (c - a)^{2} > 0

. Expand each term, group the like terms, and substitute the given value of

a^{2} + b^{2} + c^{2} = 1

. The answer will reveal itself.

Visualized Solution

Analyzing the Given Conditions for $a, b, c$

Given: $a, b, c$ are distinct positive real numbers.
Constraint: $a^{2} + b^{2} + c^{2} = 1$ .
Goal: Find the range of $ab + b c + c a$ .

The Fundamental Property: $x^{2} \geq 0$

Recall the fundamental property: For any real number $x$ , its square is always non-negative, meaning $x^{2} \geq 0$ .
We can apply this to the differences between our variables.
Consider the sum of squared differences: $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} \geq 0$ .

The Impact of $a \neq = b \neq = c$

The problem states that $a, b,$ and $c$ are distinct ( $a \neq = b \neq = c$ ).
This means $(a - b) \neq = 0$ , $(b - c) \neq = 0$ , and $(c - a) \neq = 0$ .
Therefore, the sum of their squares cannot be zero.
The inequality becomes strictly greater than zero: $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} > 0$ .

Expanding $(x - y)^{2}$

Let's expand each term using the identity $(x - y)^{2} = x^{2} + y^{2} - 2 x y$ .
Expanding $(a - b)^{2}$ gives $a^{2} + b^{2} - 2 ab$ .
Expanding $(b - c)^{2}$ gives $b^{2} + c^{2} - 2 b c$ .
Expanding $(c - a)^{2}$ gives $c^{2} + a^{2} - 2 c a$ .

The Expanded Inequality

Combining the expanded terms into our inequality:
$(a^{2} + b^{2} - 2 ab) + (b^{2} + c^{2} - 2 b c) + (c^{2} + a^{2} - 2 c a) > 0$

Grouping Like Terms

Notice that each squared term ( $a^{2}, b^{2}, c^{2}$ ) appears exactly twice.
Combine them: $2 a^{2} + 2 b^{2} + 2 c^{2} - 2 ab - 2 b c - 2 c a > 0$ .

Factoring Out the Constant $2$

Every term in the inequality has a common factor of $2$ .
Factor out the $2$ from the squared terms and the product terms separately:
$2 (a^{2} + b^{2} + c^{2}) - 2 (ab + b c + c a) > 0$ .

Substituting $a^{2} + b^{2} + c^{2} = 1$

Recall our initial constraint: $a^{2} + b^{2} + c^{2} = 1$ .
Substitute this value into our simplified inequality.
$2 (1) - 2 (ab + b c + c a) > 0$ .

Rearranging the Inequality

We have: $2 - 2 (ab + b c + c a) > 0$ .
Move the negative term to the right side of the inequality:
$2 > 2 (ab + b c + c a)$ .

Solving for $ab + b c + c a$

Divide both sides of the inequality by the positive constant $2$ .
$1 > ab + b c + c a$ .
Rewriting it from left to right: $ab + b c + c a < 1$ .

Final Conclusion: $ab + b c + c a < 1$

The value of $ab + b c + c a$ is strictly less than 1.
Key Takeaway: Using the sum of squares of differences is a powerful technique for finding bounds of symmetric algebraic expressions.

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The Sigma Insight: Linear Inequalities

Analyzing the Setup

Imagine you are standing at the edge of a mathematical landscape where

a

b

, and

c

are distinct real numbers bound by the constraint:

a^{2} + b^{2} + c^{2} = 1

Our goal is to find the range of the expression

E = ab + b c + c a

. This is a classic JEE Advanced problem that tests your intuition regarding symmetric expressions.

The Master Identity

The key lies in the identity involving the sum of squared differences:

(a - b)^{2} + (b - c)^{2} + (c - a)^{2}

This identity serves as the bridge between the sum of squares and the sum of pairwise products. Since

a

b

, and

c

are distinct, we know that

(a - b)^{2} > 0

(b - c)^{2} > 0

, and

(c - a)^{2} > 0

Consequently, their sum must be strictly greater than zero:

(a - b)^{2} + (b - c)^{2} + (c - a)^{2} > 0

The Algebraic Dance

Expanding the identity above, we obtain:

(a^{2} + b^{2} - 2 ab) + (b^{2} + c^{2} - 2 b c) + (c^{2} + a^{2} - 2 c a) > 0

Grouping the terms, we see:

2 a^{2} + 2 b^{2} + 2 c^{2} - 2 (ab + b c + c a) > 0

Factoring out the constant

2

, we arrive at:

2 (a^{2} + b^{2} + c^{2}) - 2 (ab + b c + c a) > 0

Final Calculation

Substituting the given constraint

a^{2} + b^{2} + c^{2} = 1

into our expression, we get:

2 (1) - 2 (ab + b c + c a) > 0

Rearranging the inequality, we find:

2 > 2 (ab + b c + c a)

This simplifies beautifully to the final result:

ab + b c + c a < 1

The Power of Distinctness

Many students overlook the word "distinct" in the problem statement. In competitive exams like the JEE, every word is a vital clue.

If the problem had stated that

a, b, c

are simply real numbers without the "distinct" condition, the answer would have been

ab + b c + c a \leq 1

. The moment we introduce the condition $a eq b eq c$, we are forced into the realm of strict inequalities.

This is the beauty of algebra; it is not just about manipulating symbols, but about respecting the constraints that define the system. Keep this technique in your toolkit, for it will serve you well in many future problems.

Common Traps & FAQs

Wait, why do we reject the 'equal to' case? Why is it strictly less than 1?

That is a brilliant observation! The problem explicitly states that

a, b,

and

c

are distinct. If

a = b = c

, then

(a - b)^{2}

would be zero. But because they are distinct,

(a - b)^{2}

must be strictly positive. This forces the entire sum to be strictly greater than zero, which propagates through the algebra to make our final result strictly less than 1.

Can't we just use the AM-GM inequality here?

You absolutely can! By AM-GM, we know that

a^{2} + b^{2} \geq 2 ab

b^{2} + c^{2} \geq 2 b c

, and

c^{2} + a^{2} \geq 2 c a

. Adding these gives

2 (a^{2} + b^{2} + c^{2}) \geq 2 (ab + b c + c a)

, which simplifies to

ab + b c + c a \leq a^{2} + b^{2} + c^{2} = 1

. However, AM-GM equality holds only when

a = b = c

. Since the problem specifies they are distinct, we cannot reach the equality case, hence the strict inequality.

Animated Solution for Mathematics - Basic Mathematics: If $a, b, c$ are distinct +ve real numbers and $a^{2} + b^{2} + c^{2} = 1$ then $ab + b c + c a$ is

Select Answer:

Visualized Solution

Analyzing the Given Conditions for $a, b, c$

The Fundamental Property: $x^{2} \geq 0$

The Impact of $a \neq = b \neq = c$

Expanding $(x - y)^{2}$

The Expanded Inequality

Grouping Like Terms

Factoring Out the Constant $2$

Substituting $a^{2} + b^{2} + c^{2} = 1$

Rearranging the Inequality

Solving for $ab + b c + c a$

Final Conclusion: $ab + b c + c a < 1$

The Sigma Insight: Linear Inequalities

Analyzing the Setup

The Master Identity

The Algebraic Dance

Final Calculation

The Power of Distinctness

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Animated Solution for Mathematics - Basic Mathematics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a,b,c are distinct +ve real numbers and a2+b2+c2=1 then ab+bc+ca is

Select Answer:

Visualized Solution

Analyzing the Given Conditions for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,c

The Fundamental Property: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2≥0

The Impact of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a=b=c

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x−y)2

The Expanded Inequality

Grouping Like Terms

Factoring Out the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2+b2+c2=1

Rearranging the Inequality

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ab+bc+ca

Final Conclusion: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ab+bc+ca<1

The Sigma Insight: Linear Inequalities

Analyzing the Setup

The Master Identity

The Algebraic Dance

Final Calculation

The Power of Distinctness

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Animated Solution for Mathematics - Basic Mathematics: If $a, b, c$ are distinct +ve real numbers and $a^{2} + b^{2} + c^{2} = 1$ then $ab + b c + c a$ is

Analyzing the Given Conditions for $a, b, c$

The Fundamental Property: $x^{2} \geq 0$

The Impact of $a \neq = b \neq = c$

Expanding $(x - y)^{2}$

Factoring Out the Constant $2$

Substituting $a^{2} + b^{2} + c^{2} = 1$

Solving for $ab + b c + c a$

Final Conclusion: $ab + b c + c a < 1$

If $S$ is the set of all real $x$ such that $\frac{2 x - 1}{2 x ^{3} + 3 x ^{2} + x}$ is positive, then $S$ contains

Find the set of all $x$ for which $\frac{2 x}{2 x ^{2} + 5 x + 2} > \frac{1}{x + 1}$

If $p \to (p \land \sim q)$ is false. Then the truth values of $p$ and $q$ are respectively

Let $y = \frac{( x + 1 ) ( x - 3 )}{( x - 2 )}$ . Find all the real values of $x$ for which $y$ takes real values.

The Boolean expression $(p \land \sim q) \Rightarrow (q \lor \sim p)$ is equivalent to:

Prove by mathematical induction that $\frac{( 2 n )!}{2 ^{2 n} ( n ! ) ^{2}} \leq \frac{1}{( 3 n + 1 ) ^{1/2}}$ for all positive integers $n$ .

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