Question

If a1, a2, b1, b2 are the positive real numbers then \((a_1+a_2\ \dots+a_n)(\frac{1}{a_1}+\frac{1}{a_2}\dots+\frac{1}{a_n})\) is
1. ≥ n³
2. > n²
3. ≥ n²
4. ≤ n³

Answer 1

Correct Answer - Option 3 : ≥ n²

Concept:

Cauchy - Schwarz Inequality:

a1, a2, b1, b2 are non-zero real numbers

If a1, . . . , an and b1, . . . , bn are real numbers, then

(a1a2 + b1b1 + ....anbn)2 ≤ (a12 + a22 + ...an2)(b12 + b22 + ...bn2)

= \(\displaystyle(\sum\limits_{i=1}^na_ib_i)^2≤\sum\limits_{i=1}^na_i^2\sum\limits_{i=1}^nb_i^2\)

 

Calculation:

Let,

x = \((a_1+a_2\ \dots+a_n)(\frac{1}{a_1}+\frac{1}{a_2}\dots+\frac{1}{a_n})\)

⇒ x = \([(\sqrt{a_1})^2+(\sqrt{a_2})^2\ \dots+(\sqrt{a_n})^2][\frac{1}{(\sqrt{a_1})^2}+\frac{1}{(\sqrt{a_2})^2}\dots+\frac{1}{(\sqrt{a_n})^2})\)

From the Cauchy - Schwarz Inequality, 

x ≥ \([\frac{(\sqrt{a_1})}{(\sqrt{a_1})}+\frac{(\sqrt{a_2})}{(\sqrt{a_2})}\dots+\frac{(\sqrt{a_n})}{(\sqrt{a_n})}]^2\)

⇒ x ≥ n²

Hence, required inequality

\((a_1+a_2\ \dots+a_n)(\frac{1}{a_1}+\frac{1}{a_2}\dots+\frac{1}{a_n})\ \ge\ n^2\)