Is it possible?

Algebra Level 1

There exist 3 positive numbers such that their sum is 8 and their product is 27.

Is this true?

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Skye Rzym by SKYE RZYM , 20, Indonesia

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6 solutions

Steven Yuan
Jun 15, 2017

Suppose there does exist a solution ( a , b , c ) (a, b, c) in positive real numbers to the system. Then, by the AM-GM Inequality, we expect that

a + b + c ≥ 3 a b c 3 8 ≥ 3 27 3 8 ≥ 9. \begin{aligned} a + b + c &\geq 3 \sqrt[3]{abc} \\ 8 &\geq 3 \sqrt[3]{27} \\ 8 &\geq 9. \end{aligned}

However, 8 is not greater than 9. Thus, we have a contradiction, so there are no solutions in positive real numbers to the system of equaitons.

21 Helpful 0 Interesting 0 Brilliant 1 Confused
Ei Young
Jul 2, 2017

Factoring 27 led to the prime factors 3,3,3. They do not sum to 8

7 Helpful 0 Interesting 0 Brilliant 0 Confused

The problem does not say the numbers have to be integers. There are non-integer factors of 27.

Jeremy Bohrer - 3 years, 11 months ago
Mohammad Khaza
Jul 14, 2017

Factoring 27 led to the prime factors 3,3,3.

and their sum is = 3 + 3 + 3 3+3+3 =9

They do not sum to 8

3 Helpful 0 Interesting 0 Brilliant 0 Confused

27 = 3 * 3 * 3, so for abc to be positive integers they all need to be 3, or 1 is 9, 1 is 3 and 1 is 1. Both cases don't fulfill the first statement, so it's not possible.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

But the problem does not specify a, b, c to be positive integers. How can we understand from this proof that there are no positive real numbers a, b, c satisfying the property?

Arunsoumya Basu - 3 years, 11 months ago

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Aah a my mind filled in integers instead of numbers. Need to think about it again then.

Peter van der Linden - 3 years, 11 months ago

Ah, but (a, b, c) = (3, 3, 3) is the set of positive real numbers whose product is 27 that has the smallest sum. This underlies the solution by Steven Yuan above.

Lynn Kiaer - 3 years, 11 months ago

If a product of 3 numbers is odd, none of them can be even (i.e. all the 3 are odd). A sum of 3 odd numbers is always odd. Q.E.D.

Igor Tseitkin - 3 years, 11 months ago
Data Space
Apr 6, 2018

27 is odd so we need a, b and c to be odd, but since 8 is even the sum of 3 odds can't give you an even number. So the statement cannot be true.

0 Helpful 0 Interesting 1 Brilliant 0 Confused

If we multiply x + y + z with yz, we get:

xyz + y^2z + yz^2 = 8 =>

y^2z + yz^2 = -19

Because x > 0, y > 0, z > 0 the equality above is false. So, there are no such positive numbers.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Why not x y z + y 2 z + y z 2 = 8 y z xyz+y^{2}z+yz^{2}=8yz ?

SKYE RZYM - 3 years, 11 months ago

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