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If a × b × v × d × e = 1 and a,b v,d,e are all positive real numbers; what is the minimum value of a+b+c+d+e?

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If a × b × v × d × e = 1 and a,b v,d,e are all positive real numbers; what is the minimum value of a+b+c+d+e?
posted Feb 25, 2016 by Milli Edwards

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3 Answers

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if a*b=1 and a and b re +ve (a+b) >2(get by squaring a+b). minimum should be 5 for a+b+c+d+e when all are 1

answer Feb 26, 2016 by anonymous
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a*b=1 and a and b re +ve (a+b) >2(get by squaring a+b).
minimum should be 5 for a+b+c+d+e when all are 1

answer Feb 28, 2016 by Gunjan Saraswat
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let us start with a*b=1
then b=1/a
a+b=a+(1/a)
differentiate f(a)=a+(1/a)
f '(a)=a-(1/a*a)
equate to zero
a=1,-1 since a is positive a=1, b=1.
differentiate second time
f"(a)=2/(a*a*a)
at a=1 it is positive hence minima
generalize a=b=c=d=e=1

answer Dec 13, 2016 by Kewal Panesar



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