Find $a^3 + b^3 +c^3, $ given $a+b+c=12$ and $a^3 cdot b^4 cdot c^5 = 0.1 cdot (600)^3$
$begingroup$
$a+b+c=12$ and $a^3 cdot b^4 cdot c^5 = 0.1 cdot (600)^3$.
Find $a^3 + b^3 +c^3 = ?$
My approach is to use AM-GM inequality. Is it correct?
a.m.-g.m.-inequality
edited Dec 19 '18 at 15:58
amWhy
1
asked Dec 19 '18 at 14:56
Mark 7Mark 7
95
$endgroup$
add a comment |
$begingroup$
$a+b+c=12$ and $a^3 cdot b^4 cdot c^5 = 0.1 cdot (600)^3$.
Find $a^3 + b^3 +c^3 = ?$
My approach is to use AM-GM inequality. Is it correct?
a.m.-g.m.-inequality
edited Dec 19 '18 at 15:58
amWhy
1
asked Dec 19 '18 at 14:56
Mark 7Mark 7
95
$endgroup$
$begingroup$
Instead of the lengthy title which adds nothing to the post, you should have presented your ideas in detail. Also, are there any conditions on the numbers? Positive? Integer? Rational?
$endgroup$
– A. Pongrácz
Dec 19 '18 at 15:21
add a comment |
$begingroup$
$a+b+c=12$ and $a^3 cdot b^4 cdot c^5 = 0.1 cdot (600)^3$.
Find $a^3 + b^3 +c^3 = ?$
My approach is to use AM-GM inequality. Is it correct?
a.m.-g.m.-inequality
edited Dec 19 '18 at 15:58
amWhy
1
asked Dec 19 '18 at 14:56
Mark 7Mark 7
95
$endgroup$
$a+b+c=12$ and $a^3 cdot b^4 cdot c^5 = 0.1 cdot (600)^3$.
Find $a^3 + b^3 +c^3 = ?$
My approach is to use AM-GM inequality. Is it correct?
a.m.-g.m.-inequality
a.m.-g.m.-inequality
edited Dec 19 '18 at 15:58
amWhy
1
asked Dec 19 '18 at 14:56
Mark 7Mark 7
95
edited Dec 19 '18 at 15:58
amWhy
1
asked Dec 19 '18 at 14:56
Mark 7Mark 7
95
edited Dec 19 '18 at 15:58
amWhy
1
edited Dec 19 '18 at 15:58
amWhy
1
edited Dec 19 '18 at 15:58
amWhy
1
1
asked Dec 19 '18 at 14:56
Mark 7Mark 7
95
asked Dec 19 '18 at 14:56
Mark 7Mark 7
95
asked Dec 19 '18 at 14:56
Mark 7Mark 7
95
95
$begingroup$
Instead of the lengthy title which adds nothing to the post, you should have presented your ideas in detail. Also, are there any conditions on the numbers? Positive? Integer? Rational?
$endgroup$
– A. Pongrácz
Dec 19 '18 at 15:21
add a comment |
$begingroup$
Instead of the lengthy title which adds nothing to the post, you should have presented your ideas in detail. Also, are there any conditions on the numbers? Positive? Integer? Rational?
$endgroup$
– A. Pongrácz
Dec 19 '18 at 15:21
$begingroup$
Instead of the lengthy title which adds nothing to the post, you should have presented your ideas in detail. Also, are there any conditions on the numbers? Positive? Integer? Rational?
$endgroup$
– A. Pongrácz
Dec 19 '18 at 15:21
$begingroup$
Instead of the lengthy title which adds nothing to the post, you should have presented your ideas in detail. Also, are there any conditions on the numbers? Positive? Integer? Rational?
$endgroup$
– A. Pongrácz
Dec 19 '18 at 15:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If the numbers have to be positive, then the problem can be solved using AM-GM, indeed.
Namely, consider the following 12 numbers:
$frac{a}{3}, frac{a}{3}, frac{a}{3}, frac{b}{4}, frac{b}{4}, frac{b}{4}, frac{b}{4}, frac{c}{5}, frac{c}{5}, frac{c}{5}, frac{c}{5}, frac{c}{5}$.
The AM is $1$, as $a+b+c=12$.
Thus, the GM is at most $1$, so the product of these $12$ numbers is at most $1$, which yields
$frac{a^3b^4c^5}{3^3cdot 4^4cdot 5^5}leq 1$.
As $a^3b^4c^5 = 0.1cdot 600^3 = 3^3cdot 4^4cdot 5^5$, the above inequality is sharp: it holds with equality.
Thus the AM and the GM of the given $12$ numbers coincide, which means that all $12$ numbers must be equal.
Hence, $frac{a}{3}= frac{b}{4}= frac{c}{5}$.
Using $a+b+c=12$ we obtain $a=3, b=4, c=5$.
Now you just have to sustitute this into $a^3+b^3+c^3$, and you are done.
answered Dec 19 '18 at 15:29
A. PongráczA. Pongrácz
5,9531929
$endgroup$
$begingroup$
Thank you I got it:)
$endgroup$
– Mark 7
Dec 21 '18 at 13:30
$begingroup$
I'm glad I could help. If you are happy with the answer, you should accept it.
$endgroup$
– A. Pongrácz
Dec 24 '18 at 22:08
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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$begingroup$
If the numbers have to be positive, then the problem can be solved using AM-GM, indeed.
Namely, consider the following 12 numbers:
$frac{a}{3}, frac{a}{3}, frac{a}{3}, frac{b}{4}, frac{b}{4}, frac{b}{4}, frac{b}{4}, frac{c}{5}, frac{c}{5}, frac{c}{5}, frac{c}{5}, frac{c}{5}$.
The AM is $1$, as $a+b+c=12$.
Thus, the GM is at most $1$, so the product of these $12$ numbers is at most $1$, which yields
$frac{a^3b^4c^5}{3^3cdot 4^4cdot 5^5}leq 1$.
As $a^3b^4c^5 = 0.1cdot 600^3 = 3^3cdot 4^4cdot 5^5$, the above inequality is sharp: it holds with equality.
Thus the AM and the GM of the given $12$ numbers coincide, which means that all $12$ numbers must be equal.
Hence, $frac{a}{3}= frac{b}{4}= frac{c}{5}$.
Using $a+b+c=12$ we obtain $a=3, b=4, c=5$.
Now you just have to sustitute this into $a^3+b^3+c^3$, and you are done.
answered Dec 19 '18 at 15:29
A. PongráczA. Pongrácz
5,9531929
$endgroup$
$begingroup$
Thank you I got it:)
$endgroup$
– Mark 7
Dec 21 '18 at 13:30
$begingroup$
I'm glad I could help. If you are happy with the answer, you should accept it.
$endgroup$
– A. Pongrácz
Dec 24 '18 at 22:08
add a comment |
$begingroup$
If the numbers have to be positive, then the problem can be solved using AM-GM, indeed.
Namely, consider the following 12 numbers:
$frac{a}{3}, frac{a}{3}, frac{a}{3}, frac{b}{4}, frac{b}{4}, frac{b}{4}, frac{b}{4}, frac{c}{5}, frac{c}{5}, frac{c}{5}, frac{c}{5}, frac{c}{5}$.
The AM is $1$, as $a+b+c=12$.
Thus, the GM is at most $1$, so the product of these $12$ numbers is at most $1$, which yields
$frac{a^3b^4c^5}{3^3cdot 4^4cdot 5^5}leq 1$.
As $a^3b^4c^5 = 0.1cdot 600^3 = 3^3cdot 4^4cdot 5^5$, the above inequality is sharp: it holds with equality.
Thus the AM and the GM of the given $12$ numbers coincide, which means that all $12$ numbers must be equal.
Hence, $frac{a}{3}= frac{b}{4}= frac{c}{5}$.
Using $a+b+c=12$ we obtain $a=3, b=4, c=5$.
Now you just have to sustitute this into $a^3+b^3+c^3$, and you are done.
answered Dec 19 '18 at 15:29
A. PongráczA. Pongrácz
5,9531929
$endgroup$
$begingroup$
Thank you I got it:)
$endgroup$
– Mark 7
Dec 21 '18 at 13:30
$begingroup$
I'm glad I could help. If you are happy with the answer, you should accept it.
$endgroup$
– A. Pongrácz
Dec 24 '18 at 22:08
add a comment |
$begingroup$
If the numbers have to be positive, then the problem can be solved using AM-GM, indeed.
Namely, consider the following 12 numbers:
$frac{a}{3}, frac{a}{3}, frac{a}{3}, frac{b}{4}, frac{b}{4}, frac{b}{4}, frac{b}{4}, frac{c}{5}, frac{c}{5}, frac{c}{5}, frac{c}{5}, frac{c}{5}$.
The AM is $1$, as $a+b+c=12$.
Thus, the GM is at most $1$, so the product of these $12$ numbers is at most $1$, which yields
$frac{a^3b^4c^5}{3^3cdot 4^4cdot 5^5}leq 1$.
As $a^3b^4c^5 = 0.1cdot 600^3 = 3^3cdot 4^4cdot 5^5$, the above inequality is sharp: it holds with equality.
Thus the AM and the GM of the given $12$ numbers coincide, which means that all $12$ numbers must be equal.
Hence, $frac{a}{3}= frac{b}{4}= frac{c}{5}$.
Using $a+b+c=12$ we obtain $a=3, b=4, c=5$.
Now you just have to sustitute this into $a^3+b^3+c^3$, and you are done.
answered Dec 19 '18 at 15:29
A. PongráczA. Pongrácz
5,9531929
$endgroup$
If the numbers have to be positive, then the problem can be solved using AM-GM, indeed.
Namely, consider the following 12 numbers:
$frac{a}{3}, frac{a}{3}, frac{a}{3}, frac{b}{4}, frac{b}{4}, frac{b}{4}, frac{b}{4}, frac{c}{5}, frac{c}{5}, frac{c}{5}, frac{c}{5}, frac{c}{5}$.
The AM is $1$, as $a+b+c=12$.
Thus, the GM is at most $1$, so the product of these $12$ numbers is at most $1$, which yields
$frac{a^3b^4c^5}{3^3cdot 4^4cdot 5^5}leq 1$.
As $a^3b^4c^5 = 0.1cdot 600^3 = 3^3cdot 4^4cdot 5^5$, the above inequality is sharp: it holds with equality.
Thus the AM and the GM of the given $12$ numbers coincide, which means that all $12$ numbers must be equal.
Hence, $frac{a}{3}= frac{b}{4}= frac{c}{5}$.
Using $a+b+c=12$ we obtain $a=3, b=4, c=5$.
Now you just have to sustitute this into $a^3+b^3+c^3$, and you are done.
answered Dec 19 '18 at 15:29
A. PongráczA. Pongrácz
5,9531929
answered Dec 19 '18 at 15:29
A. PongráczA. Pongrácz
5,9531929
answered Dec 19 '18 at 15:29
A. PongráczA. Pongrácz
5,9531929
answered Dec 19 '18 at 15:29
A. PongráczA. Pongrácz
5,9531929
5,9531929
$begingroup$
Thank you I got it:)
$endgroup$
– Mark 7
Dec 21 '18 at 13:30
$begingroup$
I'm glad I could help. If you are happy with the answer, you should accept it.
$endgroup$
– A. Pongrácz
Dec 24 '18 at 22:08
add a comment |
$begingroup$
Thank you I got it:)
$endgroup$
– Mark 7
Dec 21 '18 at 13:30
$begingroup$
I'm glad I could help. If you are happy with the answer, you should accept it.
$endgroup$
– A. Pongrácz
Dec 24 '18 at 22:08
$begingroup$
Thank you I got it:)
$endgroup$
– Mark 7
Dec 21 '18 at 13:30
$begingroup$
Thank you I got it:)
$endgroup$
– Mark 7
Dec 21 '18 at 13:30
$begingroup$
I'm glad I could help. If you are happy with the answer, you should accept it.
$endgroup$
– A. Pongrácz
Dec 24 '18 at 22:08
$begingroup$
I'm glad I could help. If you are happy with the answer, you should accept it.
$endgroup$
– A. Pongrácz
Dec 24 '18 at 22:08
add a comment |
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$begingroup$
Instead of the lengthy title which adds nothing to the post, you should have presented your ideas in detail. Also, are there any conditions on the numbers? Positive? Integer? Rational?
$endgroup$
– A. Pongrácz
Dec 19 '18 at 15:21