If $a_{1}, a_{2}, . . . . ., a_{n}$ and $b_{1}, b_{2}, . . . . . ., b_{n}$ be real numbers and more of the $b_{i}$ 's be zero, then prove that $(a_{1}^{2} + a_{2}^{2} + . . . . . + a_{n}^{2}) (b_{1}^{- 2} + b_{2}^{- 2} . . . . . . . . . . . . + b_{n}^{- 2}) \geq {(\frac{a_{1}}{b_{1}} + . . . . . \frac{a_{n}}{b_{n}})}^{2}$

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Solution

Applying Cauchy-Schwarz inequality to the numbers $a_{1} . . . . . . . . . . a_{n}$ ,
$b_{1}^{- 1}, . . . . . . . . . . . . . b_{n}^{- 1}$ , we have
${a_{1} (\frac{1}{b_{1}}) + . . . . . . a_{n} (\frac{1}{b_{n}})}^{2} \leq (a_{1}^{2} + . . . + a_{n}^{2}) (b_{1}^{- 2} + . . . . + b_{n}^{- 2}), o r (a_{1}^{2} + . . . . . . . . . . . . . . . a_{n}^{2}) (b_{1}^{- 2} + . . . . + b_{n}^{- 2}) \geq (\frac{a_{1}}{b_{1}} + . . . . . . . . . + \frac{a_{n}}{b_{n}})^{2}$

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Similar questions

{[{(a_{1} {+ a}_{2} + . . . . a_{n})}^{2} {(b_{1} + b_{2} + . . . . + b_{n})}^{2}]}^{1 / 2} < \sqrt{{a_{1}}^{2} + {b_{1}}^{2} +} \sqrt{{a_{2}}^{2} + {b_{2}}^{2}} + . . . . + \sqrt{{a_{n}}^{2} + {b_{n}}^{2}}

a_{r}, b_{r} (r = 1, 2, ., n)

\frac{b_{1}}{a_{1}} - \frac{b_{2}}{a_{2}} - \frac{b_{3}}{a_{3}} - \dots

p_{n} = a_{n} p_{n - 1} - b_{n} p_{n - 2}, q_{n} = a_{n} q_{n - 1} - b_{n} q_{n - 2}

(a_{1}, b_{1})

(a_{2}, b_{2})

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P (\frac{B i}{A}) = \frac{P (B i) P (\frac{A}{B i})}{\sum (B i) P (\frac{A}{B i})}

P (\frac{B i}{A}) = \frac{P (B i) P (\frac{A}{B i})}{\sum (B i) P (\frac{A}{B i})}

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