Mixing
Deducing the AM-GM inequality from a maximum argument
Clash Royale CLAN TAG#URR8PPP
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Let $f: [0, infty)^n rightarrow mathbbR$
$$f(x)=frac(x_1x_2...x_n)^frac1n+11+x_1+...+x_n$$
Show that the function achieves its global maximum at $(1,1,...,1) in mathbbR^n$ and deduce that
For all $y_1, ... , y_n+1 in [0, infty)$
$$(y_1...y_n+1)^frac1n+1leq frac1n+1 sum_i=1^n+1y_i$$
I have shown that the function's global maximum is at $(1,...,1)$, after that I get that $f(x)leq frac1n+1$, so
$$(x_1x_2...x_n)^frac1n+1leq frac1n+1(1+x_1+...+x_n)$$
I fail to see how I can show the proposition for a $n+1$. It seems like I should vary the $1$, but I fail to see how I can achieve the AM GM inequality that way. Any hints are welcome.
calculus inequality
asked Jul 18 at 16:03
B.Swan
9701619
add a comment |Â
up vote
2
down vote
favorite
Let $f: [0, infty)^n rightarrow mathbbR$
$$f(x)=frac(x_1x_2...x_n)^frac1n+11+x_1+...+x_n$$
Show that the function achieves its global maximum at $(1,1,...,1) in mathbbR^n$ and deduce that
For all $y_1, ... , y_n+1 in [0, infty)$
$$(y_1...y_n+1)^frac1n+1leq frac1n+1 sum_i=1^n+1y_i$$
I have shown that the function's global maximum is at $(1,...,1)$, after that I get that $f(x)leq frac1n+1$, so
$$(x_1x_2...x_n)^frac1n+1leq frac1n+1(1+x_1+...+x_n)$$
I fail to see how I can show the proposition for a $n+1$. It seems like I should vary the $1$, but I fail to see how I can achieve the AM GM inequality that way. Any hints are welcome.
calculus inequality
asked Jul 18 at 16:03
B.Swan
9701619
Hint: This is AM-GM in the special case where the first number (or the $(n+1)$th number, which seems to suit the indexing here better) is equal to $1$. Can you show the general case from there?
– Arthur
Jul 18 at 16:10
1
Just set $x_i = y_i / y_n+1$ ...
– Martin R
Jul 18 at 16:34
@MartinR Thanks a lot. I suspected it was something trivial (well now it is) that I wasnt seeing.
– B.Swan
Jul 18 at 16:36
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f: [0, infty)^n rightarrow mathbbR$
$$f(x)=frac(x_1x_2...x_n)^frac1n+11+x_1+...+x_n$$
Show that the function achieves its global maximum at $(1,1,...,1) in mathbbR^n$ and deduce that
For all $y_1, ... , y_n+1 in [0, infty)$
$$(y_1...y_n+1)^frac1n+1leq frac1n+1 sum_i=1^n+1y_i$$
I have shown that the function's global maximum is at $(1,...,1)$, after that I get that $f(x)leq frac1n+1$, so
$$(x_1x_2...x_n)^frac1n+1leq frac1n+1(1+x_1+...+x_n)$$
I fail to see how I can show the proposition for a $n+1$. It seems like I should vary the $1$, but I fail to see how I can achieve the AM GM inequality that way. Any hints are welcome.
calculus inequality
asked Jul 18 at 16:03
B.Swan
9701619
Let $f: [0, infty)^n rightarrow mathbbR$
$$f(x)=frac(x_1x_2...x_n)^frac1n+11+x_1+...+x_n$$
Show that the function achieves its global maximum at $(1,1,...,1) in mathbbR^n$ and deduce that
For all $y_1, ... , y_n+1 in [0, infty)$
$$(y_1...y_n+1)^frac1n+1leq frac1n+1 sum_i=1^n+1y_i$$
I have shown that the function's global maximum is at $(1,...,1)$, after that I get that $f(x)leq frac1n+1$, so
$$(x_1x_2...x_n)^frac1n+1leq frac1n+1(1+x_1+...+x_n)$$
I fail to see how I can show the proposition for a $n+1$. It seems like I should vary the $1$, but I fail to see how I can achieve the AM GM inequality that way. Any hints are welcome.
calculus inequality
asked Jul 18 at 16:03
B.Swan
9701619
asked Jul 18 at 16:03
B.Swan
9701619
asked Jul 18 at 16:03
B.Swan
9701619
asked Jul 18 at 16:03
B.Swan
9701619
9701619
Hint: This is AM-GM in the special case where the first number (or the $(n+1)$th number, which seems to suit the indexing here better) is equal to $1$. Can you show the general case from there?
– Arthur
Jul 18 at 16:10
1
Just set $x_i = y_i / y_n+1$ ...
– Martin R
Jul 18 at 16:34
@MartinR Thanks a lot. I suspected it was something trivial (well now it is) that I wasnt seeing.
– B.Swan
Jul 18 at 16:36
add a comment |Â
Hint: This is AM-GM in the special case where the first number (or the $(n+1)$th number, which seems to suit the indexing here better) is equal to $1$. Can you show the general case from there?
– Arthur
Jul 18 at 16:10
1
Just set $x_i = y_i / y_n+1$ ...
– Martin R
Jul 18 at 16:34
@MartinR Thanks a lot. I suspected it was something trivial (well now it is) that I wasnt seeing.
– B.Swan
Jul 18 at 16:36
Hint: This is AM-GM in the special case where the first number (or the $(n+1)$th number, which seems to suit the indexing here better) is equal to $1$. Can you show the general case from there?
– Arthur
Jul 18 at 16:10
Hint: This is AM-GM in the special case where the first number (or the $(n+1)$th number, which seems to suit the indexing here better) is equal to $1$. Can you show the general case from there?
– Arthur
Jul 18 at 16:10
1
1
Just set $x_i = y_i / y_n+1$ ...
– Martin R
Jul 18 at 16:34
Just set $x_i = y_i / y_n+1$ ...
– Martin R
Jul 18 at 16:34
@MartinR Thanks a lot. I suspected it was something trivial (well now it is) that I wasnt seeing.
– B.Swan
Jul 18 at 16:36
@MartinR Thanks a lot. I suspected it was something trivial (well now it is) that I wasnt seeing.
– B.Swan
Jul 18 at 16:36
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $y_1, ldots, y_n+1 in [0, +infty)$ and assume $y_n+1 ne 0$.
Then
beginalign
(y_1ldots y_n+1)^frac1n+1 &= y_n+1left(fracy_1y_n+1cdotsfracy_ny_n+1right)^frac1n+1\
&le y_n+1 frac1 + fracy_1y_n+1 + cdots + fracy_ny_n+1n+1\
&= fracy_1 + cdots + y_n + y_n+1n+1
endalign
answered Jul 18 at 16:35
mechanodroid
22.2k52041
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $y_1, ldots, y_n+1 in [0, +infty)$ and assume $y_n+1 ne 0$.
Then
beginalign
(y_1ldots y_n+1)^frac1n+1 &= y_n+1left(fracy_1y_n+1cdotsfracy_ny_n+1right)^frac1n+1\
&le y_n+1 frac1 + fracy_1y_n+1 + cdots + fracy_ny_n+1n+1\
&= fracy_1 + cdots + y_n + y_n+1n+1
endalign
answered Jul 18 at 16:35
mechanodroid
22.2k52041
add a comment |Â
up vote
2
down vote
accepted
Let $y_1, ldots, y_n+1 in [0, +infty)$ and assume $y_n+1 ne 0$.
Then
beginalign
(y_1ldots y_n+1)^frac1n+1 &= y_n+1left(fracy_1y_n+1cdotsfracy_ny_n+1right)^frac1n+1\
&le y_n+1 frac1 + fracy_1y_n+1 + cdots + fracy_ny_n+1n+1\
&= fracy_1 + cdots + y_n + y_n+1n+1
endalign
answered Jul 18 at 16:35
mechanodroid
22.2k52041
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $y_1, ldots, y_n+1 in [0, +infty)$ and assume $y_n+1 ne 0$.
Then
beginalign
(y_1ldots y_n+1)^frac1n+1 &= y_n+1left(fracy_1y_n+1cdotsfracy_ny_n+1right)^frac1n+1\
&le y_n+1 frac1 + fracy_1y_n+1 + cdots + fracy_ny_n+1n+1\
&= fracy_1 + cdots + y_n + y_n+1n+1
endalign
answered Jul 18 at 16:35
mechanodroid
22.2k52041
Let $y_1, ldots, y_n+1 in [0, +infty)$ and assume $y_n+1 ne 0$.
Then
beginalign
(y_1ldots y_n+1)^frac1n+1 &= y_n+1left(fracy_1y_n+1cdotsfracy_ny_n+1right)^frac1n+1\
&le y_n+1 frac1 + fracy_1y_n+1 + cdots + fracy_ny_n+1n+1\
&= fracy_1 + cdots + y_n + y_n+1n+1
endalign
answered Jul 18 at 16:35
mechanodroid
22.2k52041
answered Jul 18 at 16:35
mechanodroid
22.2k52041
answered Jul 18 at 16:35
mechanodroid
22.2k52041
answered Jul 18 at 16:35
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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Hint: This is AM-GM in the special case where the first number (or the $(n+1)$th number, which seems to suit the indexing here better) is equal to $1$. Can you show the general case from there?
– Arthur
Jul 18 at 16:10
1
Just set $x_i = y_i / y_n+1$ ...
– Martin R
Jul 18 at 16:34
@MartinR Thanks a lot. I suspected it was something trivial (well now it is) that I wasnt seeing.
– B.Swan
Jul 18 at 16:36