Mixing
Prove that product of 3 numbers with a fixed sum is highest when they are equal
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How to prove the fact, that product of any three numbers with a fixed sum is highest possible when they equal $sum/3$, meaning they are equal?
Example:
Let the "sum" be a constant $K$ and the "numbers" $a, b, c$.
given $K = 3$
How can I prove that $a,b,c$ make the highest product when they equal $K/3 = 1$?
Thanks in advance!
products
asked Jan 16 at 20:11
Xxx DddXxx Ddd
354
$endgroup$
add a comment |
$begingroup$
How to prove the fact, that product of any three numbers with a fixed sum is highest possible when they equal $sum/3$, meaning they are equal?
Example:
Let the "sum" be a constant $K$ and the "numbers" $a, b, c$.
given $K = 3$
How can I prove that $a,b,c$ make the highest product when they equal $K/3 = 1$?
Thanks in advance!
products
asked Jan 16 at 20:11
Xxx DddXxx Ddd
354
$endgroup$
$begingroup$
What kind of numbers do you mean?
$endgroup$
– Dr. Sonnhard Graubner
Jan 16 at 20:13
1
$begingroup$
Try this en.wikipedia.org/wiki/…
$endgroup$
– Cardioid_Ass_22
Jan 16 at 20:15
$begingroup$
@Dr.SonnhardGraubner in this case, real positive numbers
$endgroup$
– Xxx Ddd
Jan 16 at 20:17
add a comment |
$begingroup$
How to prove the fact, that product of any three numbers with a fixed sum is highest possible when they equal $sum/3$, meaning they are equal?
Example:
Let the "sum" be a constant $K$ and the "numbers" $a, b, c$.
given $K = 3$
How can I prove that $a,b,c$ make the highest product when they equal $K/3 = 1$?
Thanks in advance!
products
asked Jan 16 at 20:11
Xxx DddXxx Ddd
354
$endgroup$
How to prove the fact, that product of any three numbers with a fixed sum is highest possible when they equal $sum/3$, meaning they are equal?
Example:
Let the "sum" be a constant $K$ and the "numbers" $a, b, c$.
given $K = 3$
How can I prove that $a,b,c$ make the highest product when they equal $K/3 = 1$?
Thanks in advance!
products
products
asked Jan 16 at 20:11
Xxx DddXxx Ddd
354
asked Jan 16 at 20:11
Xxx DddXxx Ddd
354
asked Jan 16 at 20:11
Xxx DddXxx Ddd
354
asked Jan 16 at 20:11
Xxx DddXxx Ddd
354
asked Jan 16 at 20:11
Xxx DddXxx Ddd
354
354
$begingroup$
What kind of numbers do you mean?
$endgroup$
– Dr. Sonnhard Graubner
Jan 16 at 20:13
1
$begingroup$
Try this en.wikipedia.org/wiki/…
$endgroup$
– Cardioid_Ass_22
Jan 16 at 20:15
$begingroup$
@Dr.SonnhardGraubner in this case, real positive numbers
$endgroup$
– Xxx Ddd
Jan 16 at 20:17
add a comment |
$begingroup$
What kind of numbers do you mean?
$endgroup$
– Dr. Sonnhard Graubner
Jan 16 at 20:13
1
$begingroup$
Try this en.wikipedia.org/wiki/…
$endgroup$
– Cardioid_Ass_22
Jan 16 at 20:15
$begingroup$
@Dr.SonnhardGraubner in this case, real positive numbers
$endgroup$
– Xxx Ddd
Jan 16 at 20:17
$begingroup$
What kind of numbers do you mean?
$endgroup$
– Dr. Sonnhard Graubner
Jan 16 at 20:13
$begingroup$
What kind of numbers do you mean?
$endgroup$
– Dr. Sonnhard Graubner
Jan 16 at 20:13
1
1
$begingroup$
Try this en.wikipedia.org/wiki/…
$endgroup$
– Cardioid_Ass_22
Jan 16 at 20:15
$begingroup$
Try this en.wikipedia.org/wiki/…
$endgroup$
– Cardioid_Ass_22
Jan 16 at 20:15
$begingroup$
@Dr.SonnhardGraubner in this case, real positive numbers
$endgroup$
– Xxx Ddd
Jan 16 at 20:17
$begingroup$
@Dr.SonnhardGraubner in this case, real positive numbers
$endgroup$
– Xxx Ddd
Jan 16 at 20:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Set $$a+b+c=s$$ then we have $$frac{a+b+c}{3}geq sqrt[3]{abc}$$ if $$a,b,c$$ are assumed to be positive. This is the $$AM-GM$$ inequality. The equal sign holds if $$a=b=c$$
answered Jan 16 at 20:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
add a comment |
$begingroup$
It's for the same reason area and perimeter ratio is highest at a circle; its best to use the least perimeter (lowest sum of numbers),as this means we can get the most area for our perimeter. Thus it is best to use the same number thrice.
answered Jan 17 at 17:46
QuaternionQuaternion
1
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Set $$a+b+c=s$$ then we have $$frac{a+b+c}{3}geq sqrt[3]{abc}$$ if $$a,b,c$$ are assumed to be positive. This is the $$AM-GM$$ inequality. The equal sign holds if $$a=b=c$$
answered Jan 16 at 20:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
add a comment |
$begingroup$
Set $$a+b+c=s$$ then we have $$frac{a+b+c}{3}geq sqrt[3]{abc}$$ if $$a,b,c$$ are assumed to be positive. This is the $$AM-GM$$ inequality. The equal sign holds if $$a=b=c$$
answered Jan 16 at 20:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
add a comment |
$begingroup$
Set $$a+b+c=s$$ then we have $$frac{a+b+c}{3}geq sqrt[3]{abc}$$ if $$a,b,c$$ are assumed to be positive. This is the $$AM-GM$$ inequality. The equal sign holds if $$a=b=c$$
answered Jan 16 at 20:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
Set $$a+b+c=s$$ then we have $$frac{a+b+c}{3}geq sqrt[3]{abc}$$ if $$a,b,c$$ are assumed to be positive. This is the $$AM-GM$$ inequality. The equal sign holds if $$a=b=c$$
answered Jan 16 at 20:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Jan 16 at 20:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Jan 16 at 20:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Jan 16 at 20:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
76.2k42866
add a comment |
add a comment |
$begingroup$
It's for the same reason area and perimeter ratio is highest at a circle; its best to use the least perimeter (lowest sum of numbers),as this means we can get the most area for our perimeter. Thus it is best to use the same number thrice.
answered Jan 17 at 17:46
QuaternionQuaternion
1
$endgroup$
add a comment |
$begingroup$
It's for the same reason area and perimeter ratio is highest at a circle; its best to use the least perimeter (lowest sum of numbers),as this means we can get the most area for our perimeter. Thus it is best to use the same number thrice.
answered Jan 17 at 17:46
QuaternionQuaternion
1
$endgroup$
add a comment |
$begingroup$
It's for the same reason area and perimeter ratio is highest at a circle; its best to use the least perimeter (lowest sum of numbers),as this means we can get the most area for our perimeter. Thus it is best to use the same number thrice.
answered Jan 17 at 17:46
QuaternionQuaternion
1
$endgroup$
It's for the same reason area and perimeter ratio is highest at a circle; its best to use the least perimeter (lowest sum of numbers),as this means we can get the most area for our perimeter. Thus it is best to use the same number thrice.
answered Jan 17 at 17:46
QuaternionQuaternion
1
answered Jan 17 at 17:46
QuaternionQuaternion
1
answered Jan 17 at 17:46
QuaternionQuaternion
1
answered Jan 17 at 17:46
QuaternionQuaternion
1
1
add a comment |
add a comment |
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$begingroup$
What kind of numbers do you mean?
$endgroup$
– Dr. Sonnhard Graubner
Jan 16 at 20:13
1
$begingroup$
Try this en.wikipedia.org/wiki/…
$endgroup$
– Cardioid_Ass_22
Jan 16 at 20:15
$begingroup$
@Dr.SonnhardGraubner in this case, real positive numbers
$endgroup$
– Xxx Ddd
Jan 16 at 20:17