Doubt in proof of AM GM inequality by induction Method
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Proof of AM-GM inequality
So i was going through the AM-GM inequality and there is particular part i didn't quite understand. My question is IF c1 <1 and ck+1 > 1 then how are they considering
c.c2.c3...ck=1 ??
I took it as being true assuming that it is true P(k) for n=k but just because I'm denoting c = c1.ck+1 doesn't mean the value of ck+1 is gone, moreover they said that ck+1 > 1 so this could be anything then, ck+1 can be 2, 10 , 100, 500 so on...
in that case how will this c.c2.c3....cn= 1 hold true?? Please help!
inequality induction
asked Nov 18 at 10:50
RiRi
143
add a comment |
up vote
-1
down vote
favorite
Proof of AM-GM inequality
So i was going through the AM-GM inequality and there is particular part i didn't quite understand. My question is IF c1 <1 and ck+1 > 1 then how are they considering
c.c2.c3...ck=1 ??
I took it as being true assuming that it is true P(k) for n=k but just because I'm denoting c = c1.ck+1 doesn't mean the value of ck+1 is gone, moreover they said that ck+1 > 1 so this could be anything then, ck+1 can be 2, 10 , 100, 500 so on...
in that case how will this c.c2.c3....cn= 1 hold true?? Please help!
inequality induction
asked Nov 18 at 10:50
RiRi
143
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Proof of AM-GM inequality
So i was going through the AM-GM inequality and there is particular part i didn't quite understand. My question is IF c1 <1 and ck+1 > 1 then how are they considering
c.c2.c3...ck=1 ??
I took it as being true assuming that it is true P(k) for n=k but just because I'm denoting c = c1.ck+1 doesn't mean the value of ck+1 is gone, moreover they said that ck+1 > 1 so this could be anything then, ck+1 can be 2, 10 , 100, 500 so on...
in that case how will this c.c2.c3....cn= 1 hold true?? Please help!
inequality induction
asked Nov 18 at 10:50
RiRi
143
Proof of AM-GM inequality
So i was going through the AM-GM inequality and there is particular part i didn't quite understand. My question is IF c1 <1 and ck+1 > 1 then how are they considering
c.c2.c3...ck=1 ??
I took it as being true assuming that it is true P(k) for n=k but just because I'm denoting c = c1.ck+1 doesn't mean the value of ck+1 is gone, moreover they said that ck+1 > 1 so this could be anything then, ck+1 can be 2, 10 , 100, 500 so on...
in that case how will this c.c2.c3....cn= 1 hold true?? Please help!
inequality induction
inequality induction
asked Nov 18 at 10:50
RiRi
143
asked Nov 18 at 10:50
RiRi
143
asked Nov 18 at 10:50
RiRi
143
asked Nov 18 at 10:50
RiRi
143
asked Nov 18 at 10:50
RiRi
143
143
add a comment |
add a comment |
1 Answer
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Since $c_1leq1$ and $c_{n+1}geq1$, we obtain:
$$(c_1-1)left(c_{n+1}-1right)leq0$$ or
$$c_1+c_{n+1}geq1+c_1c_{n+1}.$$
Thus, by assumption of the induction (if $x_1x_2...x_n=1$ then $x_1+x_2+...+x_ngeq n$ for positives $x_i$)
we obtain:
$$c_1+c_2+...+c_n+c_{n+1}=c_1+c_{n+1}+c_2+...+c_ngeq$$
$$geq1+c_1c_{n+1}+c_2+...+c_ngeq1+left(c_1c_{n+1}right)c_2...c_n=1+n.$$
answered Nov 18 at 12:56
Michael Rozenberg
94.5k1588183
Sir thank you so much for answering, I just have a doubt. No.1) after multiplying (c1-1)(cn+1 - 1)<0 will give us 1+c1.cn+1< c1 + cn+1 right?
– RiRi
Nov 18 at 13:34
And 2nd Doubt is, I still didn't understand why c.c2.c3....cn= 1? since they already stated that cn+1 is greater than 1 and c denoted the multiplication of c1 and cn+1
– RiRi
Nov 18 at 13:35
@RiRi I fixed. The assumption of the induction is: for positives $c_i$ such that $c_1c_2...c_n=1$ we have $c_1+c_2+...+c_ngeq n.$ We can rewrite it so: For positives $x_i$ such that $x_1x_2...x_n=1$ we have $x_1+x_2+...+x_ngeq n.$ We need to prove that if $c_1c_2...c_{n+1}=1$ then $c_1+c_2+...+c_{n+1}geq n+1$, which I made. See my proof. $x_1=c_1c_{n+1},$ $x_2=c_2$,..., $x_n=c_n$ and we see that $x_1x_2...x_n=c_1c_2...c_{n+1}=1.$
– Michael Rozenberg
Nov 18 at 13:51
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Since $c_1leq1$ and $c_{n+1}geq1$, we obtain:
$$(c_1-1)left(c_{n+1}-1right)leq0$$ or
$$c_1+c_{n+1}geq1+c_1c_{n+1}.$$
Thus, by assumption of the induction (if $x_1x_2...x_n=1$ then $x_1+x_2+...+x_ngeq n$ for positives $x_i$)
we obtain:
$$c_1+c_2+...+c_n+c_{n+1}=c_1+c_{n+1}+c_2+...+c_ngeq$$
$$geq1+c_1c_{n+1}+c_2+...+c_ngeq1+left(c_1c_{n+1}right)c_2...c_n=1+n.$$
answered Nov 18 at 12:56
Michael Rozenberg
94.5k1588183
Sir thank you so much for answering, I just have a doubt. No.1) after multiplying (c1-1)(cn+1 - 1)<0 will give us 1+c1.cn+1< c1 + cn+1 right?
– RiRi
Nov 18 at 13:34
And 2nd Doubt is, I still didn't understand why c.c2.c3....cn= 1? since they already stated that cn+1 is greater than 1 and c denoted the multiplication of c1 and cn+1
– RiRi
Nov 18 at 13:35
@RiRi I fixed. The assumption of the induction is: for positives $c_i$ such that $c_1c_2...c_n=1$ we have $c_1+c_2+...+c_ngeq n.$ We can rewrite it so: For positives $x_i$ such that $x_1x_2...x_n=1$ we have $x_1+x_2+...+x_ngeq n.$ We need to prove that if $c_1c_2...c_{n+1}=1$ then $c_1+c_2+...+c_{n+1}geq n+1$, which I made. See my proof. $x_1=c_1c_{n+1},$ $x_2=c_2$,..., $x_n=c_n$ and we see that $x_1x_2...x_n=c_1c_2...c_{n+1}=1.$
– Michael Rozenberg
Nov 18 at 13:51
add a comment |
up vote
0
down vote
Since $c_1leq1$ and $c_{n+1}geq1$, we obtain:
$$(c_1-1)left(c_{n+1}-1right)leq0$$ or
$$c_1+c_{n+1}geq1+c_1c_{n+1}.$$
Thus, by assumption of the induction (if $x_1x_2...x_n=1$ then $x_1+x_2+...+x_ngeq n$ for positives $x_i$)
we obtain:
$$c_1+c_2+...+c_n+c_{n+1}=c_1+c_{n+1}+c_2+...+c_ngeq$$
$$geq1+c_1c_{n+1}+c_2+...+c_ngeq1+left(c_1c_{n+1}right)c_2...c_n=1+n.$$
answered Nov 18 at 12:56
Michael Rozenberg
94.5k1588183
Sir thank you so much for answering, I just have a doubt. No.1) after multiplying (c1-1)(cn+1 - 1)<0 will give us 1+c1.cn+1< c1 + cn+1 right?
– RiRi
Nov 18 at 13:34
And 2nd Doubt is, I still didn't understand why c.c2.c3....cn= 1? since they already stated that cn+1 is greater than 1 and c denoted the multiplication of c1 and cn+1
– RiRi
Nov 18 at 13:35
@RiRi I fixed. The assumption of the induction is: for positives $c_i$ such that $c_1c_2...c_n=1$ we have $c_1+c_2+...+c_ngeq n.$ We can rewrite it so: For positives $x_i$ such that $x_1x_2...x_n=1$ we have $x_1+x_2+...+x_ngeq n.$ We need to prove that if $c_1c_2...c_{n+1}=1$ then $c_1+c_2+...+c_{n+1}geq n+1$, which I made. See my proof. $x_1=c_1c_{n+1},$ $x_2=c_2$,..., $x_n=c_n$ and we see that $x_1x_2...x_n=c_1c_2...c_{n+1}=1.$
– Michael Rozenberg
Nov 18 at 13:51
add a comment |
up vote
0
down vote
up vote
0
down vote
Since $c_1leq1$ and $c_{n+1}geq1$, we obtain:
$$(c_1-1)left(c_{n+1}-1right)leq0$$ or
$$c_1+c_{n+1}geq1+c_1c_{n+1}.$$
Thus, by assumption of the induction (if $x_1x_2...x_n=1$ then $x_1+x_2+...+x_ngeq n$ for positives $x_i$)
we obtain:
$$c_1+c_2+...+c_n+c_{n+1}=c_1+c_{n+1}+c_2+...+c_ngeq$$
$$geq1+c_1c_{n+1}+c_2+...+c_ngeq1+left(c_1c_{n+1}right)c_2...c_n=1+n.$$
answered Nov 18 at 12:56
Michael Rozenberg
94.5k1588183
Since $c_1leq1$ and $c_{n+1}geq1$, we obtain:
$$(c_1-1)left(c_{n+1}-1right)leq0$$ or
$$c_1+c_{n+1}geq1+c_1c_{n+1}.$$
Thus, by assumption of the induction (if $x_1x_2...x_n=1$ then $x_1+x_2+...+x_ngeq n$ for positives $x_i$)
we obtain:
$$c_1+c_2+...+c_n+c_{n+1}=c_1+c_{n+1}+c_2+...+c_ngeq$$
$$geq1+c_1c_{n+1}+c_2+...+c_ngeq1+left(c_1c_{n+1}right)c_2...c_n=1+n.$$
answered Nov 18 at 12:56
Michael Rozenberg
94.5k1588183
edited Nov 18 at 13:44
answered Nov 18 at 12:56
Michael Rozenberg
94.5k1588183
answered Nov 18 at 12:56
Michael Rozenberg
94.5k1588183
answered Nov 18 at 12:56
Michael Rozenberg
94.5k1588183
94.5k1588183
Sir thank you so much for answering, I just have a doubt. No.1) after multiplying (c1-1)(cn+1 - 1)<0 will give us 1+c1.cn+1< c1 + cn+1 right?
– RiRi
Nov 18 at 13:34
And 2nd Doubt is, I still didn't understand why c.c2.c3....cn= 1? since they already stated that cn+1 is greater than 1 and c denoted the multiplication of c1 and cn+1
– RiRi
Nov 18 at 13:35
@RiRi I fixed. The assumption of the induction is: for positives $c_i$ such that $c_1c_2...c_n=1$ we have $c_1+c_2+...+c_ngeq n.$ We can rewrite it so: For positives $x_i$ such that $x_1x_2...x_n=1$ we have $x_1+x_2+...+x_ngeq n.$ We need to prove that if $c_1c_2...c_{n+1}=1$ then $c_1+c_2+...+c_{n+1}geq n+1$, which I made. See my proof. $x_1=c_1c_{n+1},$ $x_2=c_2$,..., $x_n=c_n$ and we see that $x_1x_2...x_n=c_1c_2...c_{n+1}=1.$
– Michael Rozenberg
Nov 18 at 13:51
add a comment |
Sir thank you so much for answering, I just have a doubt. No.1) after multiplying (c1-1)(cn+1 - 1)<0 will give us 1+c1.cn+1< c1 + cn+1 right?
– RiRi
Nov 18 at 13:34
And 2nd Doubt is, I still didn't understand why c.c2.c3....cn= 1? since they already stated that cn+1 is greater than 1 and c denoted the multiplication of c1 and cn+1
– RiRi
Nov 18 at 13:35
@RiRi I fixed. The assumption of the induction is: for positives $c_i$ such that $c_1c_2...c_n=1$ we have $c_1+c_2+...+c_ngeq n.$ We can rewrite it so: For positives $x_i$ such that $x_1x_2...x_n=1$ we have $x_1+x_2+...+x_ngeq n.$ We need to prove that if $c_1c_2...c_{n+1}=1$ then $c_1+c_2+...+c_{n+1}geq n+1$, which I made. See my proof. $x_1=c_1c_{n+1},$ $x_2=c_2$,..., $x_n=c_n$ and we see that $x_1x_2...x_n=c_1c_2...c_{n+1}=1.$
– Michael Rozenberg
Nov 18 at 13:51
Sir thank you so much for answering, I just have a doubt. No.1) after multiplying (c1-1)(cn+1 - 1)<0 will give us 1+c1.cn+1< c1 + cn+1 right?
– RiRi
Nov 18 at 13:34
Sir thank you so much for answering, I just have a doubt. No.1) after multiplying (c1-1)(cn+1 - 1)<0 will give us 1+c1.cn+1< c1 + cn+1 right?
– RiRi
Nov 18 at 13:34
And 2nd Doubt is, I still didn't understand why c.c2.c3....cn= 1? since they already stated that cn+1 is greater than 1 and c denoted the multiplication of c1 and cn+1
– RiRi
Nov 18 at 13:35
And 2nd Doubt is, I still didn't understand why c.c2.c3....cn= 1? since they already stated that cn+1 is greater than 1 and c denoted the multiplication of c1 and cn+1
– RiRi
Nov 18 at 13:35
@RiRi I fixed. The assumption of the induction is: for positives $c_i$ such that $c_1c_2...c_n=1$ we have $c_1+c_2+...+c_ngeq n.$ We can rewrite it so: For positives $x_i$ such that $x_1x_2...x_n=1$ we have $x_1+x_2+...+x_ngeq n.$ We need to prove that if $c_1c_2...c_{n+1}=1$ then $c_1+c_2+...+c_{n+1}geq n+1$, which I made. See my proof. $x_1=c_1c_{n+1},$ $x_2=c_2$,..., $x_n=c_n$ and we see that $x_1x_2...x_n=c_1c_2...c_{n+1}=1.$
– Michael Rozenberg
Nov 18 at 13:51
@RiRi I fixed. The assumption of the induction is: for positives $c_i$ such that $c_1c_2...c_n=1$ we have $c_1+c_2+...+c_ngeq n.$ We can rewrite it so: For positives $x_i$ such that $x_1x_2...x_n=1$ we have $x_1+x_2+...+x_ngeq n.$ We need to prove that if $c_1c_2...c_{n+1}=1$ then $c_1+c_2+...+c_{n+1}geq n+1$, which I made. See my proof. $x_1=c_1c_{n+1},$ $x_2=c_2$,..., $x_n=c_n$ and we see that $x_1x_2...x_n=c_1c_2...c_{n+1}=1.$
– Michael Rozenberg
Nov 18 at 13:51
add a comment |
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