Math, asked by Anonymous, 1 year ago

please anyone answer this question

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Answered by ashutosharyan874

Plz pardon if its wrong, I am not sure about it correctness

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ashutosharyan874: is it correct

Anonymous: yeah

ashutosharyan874: wooh, thanx

balakrishna40: it is correct. but u need proper justification.

balakrishna40: look at my solution

ashutosharyan874: yeah u r correct, but can't we just put the minimum value and get the answer instead of going so long

ashutosharyan874: anyway it is qn of mains

balakrishna40: u r correct

balakrishna40: extreme values will come when a=b=c

ashutosharyan874: I see u r a teacher,, plz give some tips for jee advanced, I have to give it on 27 th

Answered by balakrishna40

AM>=GM

$\\ a + \frac{1}{a} > = 2$

now

$\frac{ {a }^{2} + 3a + 1}{a} \\ = a + \frac{1}{a } + 3$

$> = 2 + 3$

$> = 5$

similarly

$\frac{ {b}^{2} + 3b + 1}{b} > = 5 \\ and \\ \frac{ {c}^{2} + 3c + 1}{c} > = 5$

multiply to get

$\frac{( {a}^{2} + 3a + 1)( {b}^{2} + 3b + 1)( {c}^{2} + 3c + 1) }{abc} > = 125$

minimum value is 125

Anonymous: how a+1/a >=2

balakrishna40: am>=gm

Anonymous: okk thanks

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