ABCD

Algebra Level 3

a , b , c , d > 0 a , b , c , d > 0

a + b + c + d = 12 a + b + c + d = 12

a b c d = 27 + a b + a c + a d + b c + b d + c d abcd = 27 + ab + ac + ad + bc + bd + cd

  • Find 4 3 ( a + d + 2 ( b + c ) ) \dfrac{4}{3}(a + d + 2(b + c))


The answer is 24.

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1 solution

Deepanshu Gupta
Dec 14, 2014

By Simple using AM-GM inequality , we get

( a b c d ) m a x = 81 ( a b ) m i n = 54 { (abcd) }_{ max }\quad =\quad 81\\ \\ { (\sum { ab } ) }_{ min }\quad =\quad 54 . But acc. to given Constrained in question ,

a b + 27 = a b c d \\ \sum { ab } \quad +\quad 27\quad =\quad abcd .

Equality of Variables must be attained

a = b = c = d = 3 a\quad =\quad b\quad =\quad c\quad =\quad d\quad =\quad 3 .

Ans = 24 , Q.E.D

Now This question is correct Please Kindly Review it and Remove Flag from this ! @Calvin Lin Sir

Deepanshu Gupta - 6 years, 6 months ago

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Thanks. I see that this problem has been rephrased, and 24 is now correct.

Those who previously answered 72 have been marked correct.

Calvin Lin Staff - 6 years, 6 months ago

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