IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

akanksha_d2006

the minimum value oF (A²+A+1)(B²B+1)(C²+C+1)(D²+D+1)/BACD

a.3⁴

b.3^-4

c.2⁴

d.none

DON007

HEY AKANSHA, NOTE

AM>GM

THEREFORE WHEN U WILL DIVIDE SEPERATELY U WILL GET

( A+ 1/A +1)( B+1/B+1)(C+1/C+1)(D+1/D+1)

AND A

by using AM>GM

A+1/A>2, B+1/B>2.....SO ON

THEREFORE

putting this we get .

(2+1)(2+1)(2+1)(2+1) = 3⁴

^{ans is A}

^{RIGHT ......................}

iitkgp_bipin

Don has done it perfectly.

Since a + 1/a > 2
Similarly b + 1/b > 2, c + 1/c > 2 , d + 1/d > 2

Hence the expression > (2+1)⁴
i.e. expression > 3⁴

Best Wishes

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