If a 0, b 0 and c 0, prove that (a + b + c) (1/a + 1/b + 1/c) ≥ 9.

Q: Wait, why can't I just expand the brackets and simplify?

Expanding the brackets gives you $1 + \frac{a}{b} + \frac{a}{c} + \frac{b}{a} + 1 + \frac{b}{c} + \frac{c}{a} + \frac{c}{b} + 1$, which simplifies to $3 + (\frac{a}{b} + \frac{b}{a}) + (\frac{a}{c} + \frac{c}{a}) + (\frac{b}{c} + \frac{c}{b})$. While this is a valid path, you would then need to apply the AM-GM inequality to each pair (e.g., $\frac{a}{b} + \frac{b}{a} \ge 2$). It works, but using the AM-HM inequality directly is much faster and more elegant!

Q: Does this inequality hold if $a, b, c$ are negative?

Great question! No, it does not. The AM-HM inequality is strictly defined for positive real numbers. If $a, b, c$ were negative, the signs of the terms would change, and the inequality could flip or become invalid. Always check your constraints—the fact that $a, b, c > 0$ is the foundation of this entire proof.

Q: When exactly does the equality hold?

The equality holds when the Arithmetic Mean equals the Harmonic Mean. In the AM-HM inequality, this happens if and only if all the numbers are equal. So, for this problem, the minimum value of $9$ is achieved when $a = b = c$.

Question

If $a > 0, b > 0$ and $c > 0$, prove that $(a + b + c) \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \ge 9$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path of mathematics. Today, we are going to unravel a classic, a problem that serves as a rite of passage for every student preparing for the JEE Advanced.

We are tasked with proving that for any positive real numbers $a, b, c$, the product of their sum and the sum of their reciprocals is at least nine:

$$(a + b + c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ight) \ge 9$$

At first glance, this might look like a daunting algebraic mess. However, by observing the symmetry and the interplay between the numbers and their reciprocals, we see a beautiful demonstration of the harmony between different types of means.

### The Master Key: AM-HM Inequality

Whenever you encounter a problem involving the sum of variables and the sum of their reciprocals, I want a specific bell to ring in your mind: the **AM-HM inequality**.

The Arithmetic Mean (AM) and the Harmonic Mean (HM) are two fundamental ways to describe the 'average' of a set of numbers. The AM-HM inequality states that for any set of positive real numbers, the Arithmetic Mean is always greater than or equal to the Harmonic Mean.

Mathematically, this is expressed as:

$$AM \ge HM$$

This is our master key. It is the tool that will unlock the door to the solution without us having to trudge through tedious expansion.

### Defining Our Tools

Let us define our tools precisely. For our three positive numbers $a, b, c$, the Arithmetic Mean is the sum of the numbers divided by the count:

$$AM = \frac{a + b + c}{3}$$

Now, let us define the Harmonic Mean. The Harmonic Mean is the reciprocal of the arithmetic mean of the reciprocals. For three numbers, this simplifies to:

$$HM = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$

Look closely at this expression. The denominator is exactly the second bracket in our target inequality. The structure is already beginning to reveal itself.

### The Synthesis

Now, we bring these two concepts together. We know that $AM \ge HM$. Substituting our definitions, we get:

$$\frac{a + b + c}{3} \ge \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$

This is the heart of the problem. We have successfully translated the verbal statement of the inequality into a rigorous mathematical relationship.

Since we are given that $a, b, c > 0$, we know that the sum of their reciprocals is also positive. This is crucial because it allows us to cross-multiply without worrying about flipping the inequality sign.

We multiply both sides by $3$ and by the sum of the reciprocals:

$$(a + b + c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ight) \ge 3 	imes 3$$

And there it is. The right side simplifies to $9$, and we have arrived at our destination:

$$(a + b + c) \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} ight) \ge 9$$

### A Final Reflection

Isn't it satisfying? We didn't need to expand a single bracket or perform complex algebraic gymnastics. By recognizing the underlying structure and applying the right theorem, the solution unfolded naturally.

Remember, the equality condition—where the expression equals exactly $9$—occurs only when $a = b = c$. This is the point of perfect balance. Keep this tool in your arsenal, and the next time you see a sum of variables and their reciprocals, you will know exactly how to handle it.

Animated Solution for Mathematics - Sequence and Series: If $a > 0, b > 0$ and $c > 0$ , prove that $(a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9$ .

Visualized Solution

Introduction to the Problem

The $A M \geq H M$ Tool

Defining Arithmetic Mean ( $A M$ )

Defining Harmonic Mean ( $H M$ )

Applying $A M \geq H M$ Inequality

Final Rearrangement

Conclusion and Key Takeaway

The Sigma Insight: Relation Between A.M., G.M., and H.M.

Analyzing the Setup

The Master Key

Defining Our Tools

The Synthesis

A Final Reflection

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Animated Solution for Mathematics - Sequence and Series: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a>0,b>0 and c>0, prove that (a+b+c)(a1​+b1​+c1​)≥9.

Visualized Solution

Introduction to the Problem

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AM≥HM Tool

Defining Arithmetic Mean ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AM)

Defining Harmonic Mean ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } HM)

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AM≥HM Inequality