If a > 0, b > 0 and c > 0, prove that...

MathematicsRelation Between A.M., G.M., and H.M.JEE Advanced 1984Easy

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 (English)

Introduction to the Problem

Given: $a, b, c > 0$
To Prove: $(a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9$

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A M \geq H M Tool

Concept: For positive real numbers, Arithmetic Mean (AM) $\geq$ Harmonic Mean (HM).
This inequality holds for any $n$ positive numbers.

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

For three numbers $a, b, c$ :
$A M = \frac{a + b + c}{3}$

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

For three numbers $a, b, c$ :
$H M = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$

Applying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A M \geq H M Inequality

Substitute $A M$ and $H M$ into the inequality:
$\frac{a + b + c}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$

Final Rearrangement

Multiply both sides by $3$ and by $(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$ :
$(a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 3 \times 3$
$(a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9$

Conclusion and Key Takeaway

Final Result: $(a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9$
Equality Condition: $a = b = c$
Generalization: For $n$ positive numbers, $(\sum a_{i}) (\sum \frac{1}{a _{i}}) \geq n^{2}$

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