Inspired by Lakshya Sinha

Algebra Level 4

If a 1 3 + a 2 3 + … + a 8 3 ≥ 8 a_1^3+a_2^3+\ldots+a_8^3\geq{8} for non-negative real a k a_k , what is the minimal value of a 1 + a 2 + … + a 8 a_1+a_2+\ldots+a_8 ?

Bonus : What is the minimum (or infimum) of a 1 + a 2 + ⋯ + a 8 a_1+a_2+\cdots+a_8 if the a k a_k are arbitrary real numbers?

Inspiration .


The answer is 2.

Otto Bretscher by Otto Bretscher , 65, USA

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2 solutions

Otto Bretscher
Sep 27, 2015

Since a k ≥ 0 a_k\geq{0} , we have ( a 1 + . . . + a 8 ) 3 ≥ a 1 3 + . . . + a 8 3 ≥ 8 (a_1+...+a_8)^3\geq{a_1^3+...+a_8^3}\geq{8} so a 1 + . . . + a 8 ≥ 2 a_1+...+a_8\geq{2} The value 2 \boxed{2} is attained if we let a 1 = 2 a_1=2 and a k = 0 a_k=0 for k > 1 k>1 .

I will let the interested reader enjoy the bonus question.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Nice approach with ignoring of additional non-negative terms.

I think the answer to bonus question is also 2 as we can choose any real number . It can be near 0 like 0.00000000000000000000000000000000000000000000000000000000000003 etc and the a1 is still 2

0 Helpful 0 Interesting 0 Brilliant 0 Confused

What about making a 1 = 3 a_1=3 and a k = − 1 a_k=-1 for k = 2 , … , 8 k=2, \ldots, 8 ?

Otto Bretscher - 5 years, 8 months ago

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