Application of weighted AM-GM
$begingroup$
Weighted AM-GM is usually stated as follows:
given the non-negative reals $a_1,a_2,dots,a_n$ and $omega_1,omega_2,dots,omega_nge 0$ with $omega_1+omega_2+dots+omega_n=1$ we have:$$omega_1 a_1+omega_2 a_2+dots+omega_n a_nge a_1^{omega_1} a_2^{omega_2}dots a_n^{omega_n}$$
Now, I was reading Mildorf's introduction to inequalities, he applies weighted AM-GM as follows:$$x_1+frac{x_2^2}{2}+frac{x_3^3}{3}+dots+frac{x_n^n}{n}ge (1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n})cdot:sqrt[1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n}]{x_1x_2x_3...x_n}$$
How does he apply weighted AM-GM to obtain that?
inequality a.m.-g.m.-inequality
edited Dec 7 '18 at 14:54
user10354138
7,3872925
asked Dec 7 '18 at 14:34
Spasoje DurovicSpasoje Durovic
35010
$endgroup$
add a comment |
$begingroup$
Weighted AM-GM is usually stated as follows:
given the non-negative reals $a_1,a_2,dots,a_n$ and $omega_1,omega_2,dots,omega_nge 0$ with $omega_1+omega_2+dots+omega_n=1$ we have:$$omega_1 a_1+omega_2 a_2+dots+omega_n a_nge a_1^{omega_1} a_2^{omega_2}dots a_n^{omega_n}$$
Now, I was reading Mildorf's introduction to inequalities, he applies weighted AM-GM as follows:$$x_1+frac{x_2^2}{2}+frac{x_3^3}{3}+dots+frac{x_n^n}{n}ge (1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n})cdot:sqrt[1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n}]{x_1x_2x_3...x_n}$$
How does he apply weighted AM-GM to obtain that?
inequality a.m.-g.m.-inequality
edited Dec 7 '18 at 14:54
user10354138
7,3872925
asked Dec 7 '18 at 14:34
Spasoje DurovicSpasoje Durovic
35010
$endgroup$
$begingroup$
Let $H_n:=sum_{i=1}^n 1/i$. Is your right-hand side meant to be $(H_n^{H_n})(prod_i x_i)^{1/2}$, or $H_n(prod_i x^i)^{1/H_n}$?
$endgroup$
– J.G.
Dec 7 '18 at 14:44
$begingroup$
@J.G. The second one, sorry for the unclear LaTeX, it's $H_ncdot sqrt[H_n]{x_1x_2x_3...x_n}$ i.e. the $H_n$-th root of the product
$endgroup$
– Spasoje Durovic
Dec 7 '18 at 14:52
add a comment |
$begingroup$
Weighted AM-GM is usually stated as follows:
given the non-negative reals $a_1,a_2,dots,a_n$ and $omega_1,omega_2,dots,omega_nge 0$ with $omega_1+omega_2+dots+omega_n=1$ we have:$$omega_1 a_1+omega_2 a_2+dots+omega_n a_nge a_1^{omega_1} a_2^{omega_2}dots a_n^{omega_n}$$
Now, I was reading Mildorf's introduction to inequalities, he applies weighted AM-GM as follows:$$x_1+frac{x_2^2}{2}+frac{x_3^3}{3}+dots+frac{x_n^n}{n}ge (1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n})cdot:sqrt[1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n}]{x_1x_2x_3...x_n}$$
How does he apply weighted AM-GM to obtain that?
inequality a.m.-g.m.-inequality
edited Dec 7 '18 at 14:54
user10354138
7,3872925
asked Dec 7 '18 at 14:34
Spasoje DurovicSpasoje Durovic
35010
$endgroup$
Weighted AM-GM is usually stated as follows:
given the non-negative reals $a_1,a_2,dots,a_n$ and $omega_1,omega_2,dots,omega_nge 0$ with $omega_1+omega_2+dots+omega_n=1$ we have:$$omega_1 a_1+omega_2 a_2+dots+omega_n a_nge a_1^{omega_1} a_2^{omega_2}dots a_n^{omega_n}$$
Now, I was reading Mildorf's introduction to inequalities, he applies weighted AM-GM as follows:$$x_1+frac{x_2^2}{2}+frac{x_3^3}{3}+dots+frac{x_n^n}{n}ge (1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n})cdot:sqrt[1+frac{1}{2}+frac{1}{3}+dots+frac{1}{n}]{x_1x_2x_3...x_n}$$
How does he apply weighted AM-GM to obtain that?
inequality a.m.-g.m.-inequality
inequality a.m.-g.m.-inequality
edited Dec 7 '18 at 14:54
user10354138
7,3872925
asked Dec 7 '18 at 14:34
Spasoje DurovicSpasoje Durovic
35010
edited Dec 7 '18 at 14:54
user10354138
7,3872925
asked Dec 7 '18 at 14:34
Spasoje DurovicSpasoje Durovic
35010
edited Dec 7 '18 at 14:54
user10354138
7,3872925
edited Dec 7 '18 at 14:54
user10354138
7,3872925
edited Dec 7 '18 at 14:54
user10354138
7,3872925
7,3872925
asked Dec 7 '18 at 14:34
Spasoje DurovicSpasoje Durovic
35010
asked Dec 7 '18 at 14:34
Spasoje DurovicSpasoje Durovic
35010
asked Dec 7 '18 at 14:34
Spasoje DurovicSpasoje Durovic
35010
35010
$begingroup$
Let $H_n:=sum_{i=1}^n 1/i$. Is your right-hand side meant to be $(H_n^{H_n})(prod_i x_i)^{1/2}$, or $H_n(prod_i x^i)^{1/H_n}$?
$endgroup$
– J.G.
Dec 7 '18 at 14:44
$begingroup$
@J.G. The second one, sorry for the unclear LaTeX, it's $H_ncdot sqrt[H_n]{x_1x_2x_3...x_n}$ i.e. the $H_n$-th root of the product
$endgroup$
– Spasoje Durovic
Dec 7 '18 at 14:52
add a comment |
$begingroup$
Let $H_n:=sum_{i=1}^n 1/i$. Is your right-hand side meant to be $(H_n^{H_n})(prod_i x_i)^{1/2}$, or $H_n(prod_i x^i)^{1/H_n}$?
$endgroup$
– J.G.
Dec 7 '18 at 14:44
$begingroup$
@J.G. The second one, sorry for the unclear LaTeX, it's $H_ncdot sqrt[H_n]{x_1x_2x_3...x_n}$ i.e. the $H_n$-th root of the product
$endgroup$
– Spasoje Durovic
Dec 7 '18 at 14:52
$begingroup$
Let $H_n:=sum_{i=1}^n 1/i$. Is your right-hand side meant to be $(H_n^{H_n})(prod_i x_i)^{1/2}$, or $H_n(prod_i x^i)^{1/H_n}$?
$endgroup$
– J.G.
Dec 7 '18 at 14:44
$begingroup$
Let $H_n:=sum_{i=1}^n 1/i$. Is your right-hand side meant to be $(H_n^{H_n})(prod_i x_i)^{1/2}$, or $H_n(prod_i x^i)^{1/H_n}$?
$endgroup$
– J.G.
Dec 7 '18 at 14:44
$begingroup$
@J.G. The second one, sorry for the unclear LaTeX, it's $H_ncdot sqrt[H_n]{x_1x_2x_3...x_n}$ i.e. the $H_n$-th root of the product
$endgroup$
– Spasoje Durovic
Dec 7 '18 at 14:52
$begingroup$
@J.G. The second one, sorry for the unclear LaTeX, it's $H_ncdot sqrt[H_n]{x_1x_2x_3...x_n}$ i.e. the $H_n$-th root of the product
$endgroup$
– Spasoje Durovic
Dec 7 '18 at 14:52
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
A proof written while discovering the solution:
Define $H_n:=sum_{i=1}^nfrac{1}{i}$ so the desired result is $sum_idfrac{x_i^i}{iH_n}geprod_ix_i^{1/H_n}$. Choosing weights $omega_i$, the right-hand side is the weighted GM of the $x_i^{1/(omega_i H_n)}$. Choose $omega_i=frac{1}{iH_n}$, so $sum_iomega_i=1$; the weighted GM of the $x_i^{1/omega_i H_n}$ is then $sum_idfrac{x_i^i}{iH_n}$ as required.
A neater version of the proof that doesn't show the discovery strategy:
With $omega_i=frac{1}{iH_n}$ and $x_i^i$ instead of $x_i$ throughout, $sum_idfrac{x_i^i}{iH_n}geprod_ix_i^{1/H_n}$ immediately follows; then multiply by $H_n$.
answered Dec 7 '18 at 14:53
J.G.J.G.
25.2k22539
$endgroup$
add a comment |
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1 Answer
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$begingroup$
A proof written while discovering the solution:
Define $H_n:=sum_{i=1}^nfrac{1}{i}$ so the desired result is $sum_idfrac{x_i^i}{iH_n}geprod_ix_i^{1/H_n}$. Choosing weights $omega_i$, the right-hand side is the weighted GM of the $x_i^{1/(omega_i H_n)}$. Choose $omega_i=frac{1}{iH_n}$, so $sum_iomega_i=1$; the weighted GM of the $x_i^{1/omega_i H_n}$ is then $sum_idfrac{x_i^i}{iH_n}$ as required.
A neater version of the proof that doesn't show the discovery strategy:
With $omega_i=frac{1}{iH_n}$ and $x_i^i$ instead of $x_i$ throughout, $sum_idfrac{x_i^i}{iH_n}geprod_ix_i^{1/H_n}$ immediately follows; then multiply by $H_n$.
answered Dec 7 '18 at 14:53
J.G.J.G.
25.2k22539
$endgroup$
add a comment |
$begingroup$
A proof written while discovering the solution:
Define $H_n:=sum_{i=1}^nfrac{1}{i}$ so the desired result is $sum_idfrac{x_i^i}{iH_n}geprod_ix_i^{1/H_n}$. Choosing weights $omega_i$, the right-hand side is the weighted GM of the $x_i^{1/(omega_i H_n)}$. Choose $omega_i=frac{1}{iH_n}$, so $sum_iomega_i=1$; the weighted GM of the $x_i^{1/omega_i H_n}$ is then $sum_idfrac{x_i^i}{iH_n}$ as required.
A neater version of the proof that doesn't show the discovery strategy:
With $omega_i=frac{1}{iH_n}$ and $x_i^i$ instead of $x_i$ throughout, $sum_idfrac{x_i^i}{iH_n}geprod_ix_i^{1/H_n}$ immediately follows; then multiply by $H_n$.
answered Dec 7 '18 at 14:53
J.G.J.G.
25.2k22539
$endgroup$
add a comment |
$begingroup$
A proof written while discovering the solution:
Define $H_n:=sum_{i=1}^nfrac{1}{i}$ so the desired result is $sum_idfrac{x_i^i}{iH_n}geprod_ix_i^{1/H_n}$. Choosing weights $omega_i$, the right-hand side is the weighted GM of the $x_i^{1/(omega_i H_n)}$. Choose $omega_i=frac{1}{iH_n}$, so $sum_iomega_i=1$; the weighted GM of the $x_i^{1/omega_i H_n}$ is then $sum_idfrac{x_i^i}{iH_n}$ as required.
A neater version of the proof that doesn't show the discovery strategy:
With $omega_i=frac{1}{iH_n}$ and $x_i^i$ instead of $x_i$ throughout, $sum_idfrac{x_i^i}{iH_n}geprod_ix_i^{1/H_n}$ immediately follows; then multiply by $H_n$.
answered Dec 7 '18 at 14:53
J.G.J.G.
25.2k22539
$endgroup$
A proof written while discovering the solution:
Define $H_n:=sum_{i=1}^nfrac{1}{i}$ so the desired result is $sum_idfrac{x_i^i}{iH_n}geprod_ix_i^{1/H_n}$. Choosing weights $omega_i$, the right-hand side is the weighted GM of the $x_i^{1/(omega_i H_n)}$. Choose $omega_i=frac{1}{iH_n}$, so $sum_iomega_i=1$; the weighted GM of the $x_i^{1/omega_i H_n}$ is then $sum_idfrac{x_i^i}{iH_n}$ as required.
A neater version of the proof that doesn't show the discovery strategy:
With $omega_i=frac{1}{iH_n}$ and $x_i^i$ instead of $x_i$ throughout, $sum_idfrac{x_i^i}{iH_n}geprod_ix_i^{1/H_n}$ immediately follows; then multiply by $H_n$.
answered Dec 7 '18 at 14:53
J.G.J.G.
25.2k22539
answered Dec 7 '18 at 14:53
J.G.J.G.
25.2k22539
answered Dec 7 '18 at 14:53
J.G.J.G.
25.2k22539
answered Dec 7 '18 at 14:53
J.G.J.G.
25.2k22539
25.2k22539
add a comment |
add a comment |
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$begingroup$
Let $H_n:=sum_{i=1}^n 1/i$. Is your right-hand side meant to be $(H_n^{H_n})(prod_i x_i)^{1/2}$, or $H_n(prod_i x^i)^{1/H_n}$?
$endgroup$
– J.G.
Dec 7 '18 at 14:44
$begingroup$
@J.G. The second one, sorry for the unclear LaTeX, it's $H_ncdot sqrt[H_n]{x_1x_2x_3...x_n}$ i.e. the $H_n$-th root of the product
$endgroup$
– Spasoje Durovic
Dec 7 '18 at 14:52