Mixing
Prove $x_1^2 + x_2^3 + … + x_{n - 1}^n + frac{1}{x_1^2 x_2^3 … x_{n - 1}^n} geq n + (x_1 - 1)^2 + 2(x_2 -...
$begingroup$
Prove that if $x_1, ..., x_{n-1}$ are positive numbers and $n geq 2$, than the following inequality holds:
$x_1^2 + x_2^3 + ... + x_{n - 1}^n + frac{1}{x_1^2 x_2^3 ... x_{n - 1}^n} geq n + (x_1 - 1)^2 + 2(x_2 - 1)^2 + ... + (n - 1)(x_{n - 1} - 1)^2$.
Firstly, I tried to prove this inequality by induction, but it didn't work. Then I tried using the AM-GM inequality, and finally some substitutions, but none of them lead me to a nicer form.
Do you have any suggestions for this inequality?
inequality induction a.m.-g.m.-inequality cauchy-schwarz-inequality
asked Dec 31 '18 at 10:46
SandelSandel
1765
$endgroup$
add a comment |
$begingroup$
Prove that if $x_1, ..., x_{n-1}$ are positive numbers and $n geq 2$, than the following inequality holds:
$x_1^2 + x_2^3 + ... + x_{n - 1}^n + frac{1}{x_1^2 x_2^3 ... x_{n - 1}^n} geq n + (x_1 - 1)^2 + 2(x_2 - 1)^2 + ... + (n - 1)(x_{n - 1} - 1)^2$.
Firstly, I tried to prove this inequality by induction, but it didn't work. Then I tried using the AM-GM inequality, and finally some substitutions, but none of them lead me to a nicer form.
Do you have any suggestions for this inequality?
inequality induction a.m.-g.m.-inequality cauchy-schwarz-inequality
asked Dec 31 '18 at 10:46
SandelSandel
1765
$endgroup$
$begingroup$
And what about $n$? Is $$ngeq 1$$?
$endgroup$
– Dr. Sonnhard Graubner
Dec 31 '18 at 10:51
$begingroup$
Could you show your work, having proceeded as you mention?
$endgroup$
– amWhy
Dec 31 '18 at 21:52
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
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– Carl Mummert
Jan 11 at 13:55
add a comment |
$begingroup$
Prove that if $x_1, ..., x_{n-1}$ are positive numbers and $n geq 2$, than the following inequality holds:
$x_1^2 + x_2^3 + ... + x_{n - 1}^n + frac{1}{x_1^2 x_2^3 ... x_{n - 1}^n} geq n + (x_1 - 1)^2 + 2(x_2 - 1)^2 + ... + (n - 1)(x_{n - 1} - 1)^2$.
Firstly, I tried to prove this inequality by induction, but it didn't work. Then I tried using the AM-GM inequality, and finally some substitutions, but none of them lead me to a nicer form.
Do you have any suggestions for this inequality?
inequality induction a.m.-g.m.-inequality cauchy-schwarz-inequality
asked Dec 31 '18 at 10:46
SandelSandel
1765
$endgroup$
Prove that if $x_1, ..., x_{n-1}$ are positive numbers and $n geq 2$, than the following inequality holds:
$x_1^2 + x_2^3 + ... + x_{n - 1}^n + frac{1}{x_1^2 x_2^3 ... x_{n - 1}^n} geq n + (x_1 - 1)^2 + 2(x_2 - 1)^2 + ... + (n - 1)(x_{n - 1} - 1)^2$.
Firstly, I tried to prove this inequality by induction, but it didn't work. Then I tried using the AM-GM inequality, and finally some substitutions, but none of them lead me to a nicer form.
Do you have any suggestions for this inequality?
inequality induction a.m.-g.m.-inequality cauchy-schwarz-inequality
inequality induction a.m.-g.m.-inequality cauchy-schwarz-inequality
asked Dec 31 '18 at 10:46
SandelSandel
1765
asked Dec 31 '18 at 10:46
SandelSandel
1765
edited Dec 31 '18 at 10:54
Sandel
asked Dec 31 '18 at 10:46
SandelSandel
1765
asked Dec 31 '18 at 10:46
SandelSandel
1765
asked Dec 31 '18 at 10:46
SandelSandel
1765
1765
$begingroup$
And what about $n$? Is $$ngeq 1$$?
$endgroup$
– Dr. Sonnhard Graubner
Dec 31 '18 at 10:51
$begingroup$
Could you show your work, having proceeded as you mention?
$endgroup$
– amWhy
Dec 31 '18 at 21:52
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Jan 11 at 13:55
add a comment |
$begingroup$
And what about $n$? Is $$ngeq 1$$?
$endgroup$
– Dr. Sonnhard Graubner
Dec 31 '18 at 10:51
$begingroup$
Could you show your work, having proceeded as you mention?
$endgroup$
– amWhy
Dec 31 '18 at 21:52
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Jan 11 at 13:55
$begingroup$
And what about $n$? Is $$ngeq 1$$?
$endgroup$
– Dr. Sonnhard Graubner
Dec 31 '18 at 10:51
$begingroup$
And what about $n$? Is $$ngeq 1$$?
$endgroup$
– Dr. Sonnhard Graubner
Dec 31 '18 at 10:51
$begingroup$
Could you show your work, having proceeded as you mention?
$endgroup$
– amWhy
Dec 31 '18 at 21:52
$begingroup$
Could you show your work, having proceeded as you mention?
$endgroup$
– amWhy
Dec 31 '18 at 21:52
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Jan 11 at 13:55
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Jan 11 at 13:55
add a comment |
1 Answer
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For all $ngeq2$ we need to prove that
$$sum_{k=1}^{n-1}(x_k^{k+1}-k(x_k-1)^2)+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geq n$$ or
$$sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+2kx_k)+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n-1)}{2}+n$$ or
$$sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+(k-1)x_k)+sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2},$$
which is true by AM-GM:
$$x_k^{k+1}-kx_k^2+(k-1)x_k=k_k(x_k^k+(k-1)cdot1-kx_k)geq0$$ and
$$sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2}left(prodlimits_{k=1}^{n-1}x_k^{k+1}cdotfrac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}right)^{frac{2}{n(n+1)}}=frac{n(n+1)}{2}.$$
answered Dec 31 '18 at 11:43
Michael RozenbergMichael Rozenberg
98.6k1590189
$endgroup$
$begingroup$
Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
$endgroup$
– Sandel
Dec 31 '18 at 17:49
$begingroup$
Happy New Year!
$endgroup$
– Michael Rozenberg
Dec 31 '18 at 17:50
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For all $ngeq2$ we need to prove that
$$sum_{k=1}^{n-1}(x_k^{k+1}-k(x_k-1)^2)+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geq n$$ or
$$sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+2kx_k)+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n-1)}{2}+n$$ or
$$sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+(k-1)x_k)+sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2},$$
which is true by AM-GM:
$$x_k^{k+1}-kx_k^2+(k-1)x_k=k_k(x_k^k+(k-1)cdot1-kx_k)geq0$$ and
$$sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2}left(prodlimits_{k=1}^{n-1}x_k^{k+1}cdotfrac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}right)^{frac{2}{n(n+1)}}=frac{n(n+1)}{2}.$$
answered Dec 31 '18 at 11:43
Michael RozenbergMichael Rozenberg
98.6k1590189
$endgroup$
$begingroup$
Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
$endgroup$
– Sandel
Dec 31 '18 at 17:49
$begingroup$
Happy New Year!
$endgroup$
– Michael Rozenberg
Dec 31 '18 at 17:50
add a comment |
$begingroup$
For all $ngeq2$ we need to prove that
$$sum_{k=1}^{n-1}(x_k^{k+1}-k(x_k-1)^2)+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geq n$$ or
$$sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+2kx_k)+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n-1)}{2}+n$$ or
$$sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+(k-1)x_k)+sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2},$$
which is true by AM-GM:
$$x_k^{k+1}-kx_k^2+(k-1)x_k=k_k(x_k^k+(k-1)cdot1-kx_k)geq0$$ and
$$sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2}left(prodlimits_{k=1}^{n-1}x_k^{k+1}cdotfrac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}right)^{frac{2}{n(n+1)}}=frac{n(n+1)}{2}.$$
answered Dec 31 '18 at 11:43
Michael RozenbergMichael Rozenberg
98.6k1590189
$endgroup$
$begingroup$
Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
$endgroup$
– Sandel
Dec 31 '18 at 17:49
$begingroup$
Happy New Year!
$endgroup$
– Michael Rozenberg
Dec 31 '18 at 17:50
add a comment |
$begingroup$
For all $ngeq2$ we need to prove that
$$sum_{k=1}^{n-1}(x_k^{k+1}-k(x_k-1)^2)+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geq n$$ or
$$sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+2kx_k)+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n-1)}{2}+n$$ or
$$sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+(k-1)x_k)+sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2},$$
which is true by AM-GM:
$$x_k^{k+1}-kx_k^2+(k-1)x_k=k_k(x_k^k+(k-1)cdot1-kx_k)geq0$$ and
$$sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2}left(prodlimits_{k=1}^{n-1}x_k^{k+1}cdotfrac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}right)^{frac{2}{n(n+1)}}=frac{n(n+1)}{2}.$$
answered Dec 31 '18 at 11:43
Michael RozenbergMichael Rozenberg
98.6k1590189
$endgroup$
For all $ngeq2$ we need to prove that
$$sum_{k=1}^{n-1}(x_k^{k+1}-k(x_k-1)^2)+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geq n$$ or
$$sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+2kx_k)+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n-1)}{2}+n$$ or
$$sum_{k=1}^{n-1}(x_k^{k+1}-kx_k^2+(k-1)x_k)+sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2},$$
which is true by AM-GM:
$$x_k^{k+1}-kx_k^2+(k-1)x_k=k_k(x_k^k+(k-1)cdot1-kx_k)geq0$$ and
$$sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2}left(prodlimits_{k=1}^{n-1}x_k^{k+1}cdotfrac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}right)^{frac{2}{n(n+1)}}=frac{n(n+1)}{2}.$$
answered Dec 31 '18 at 11:43
Michael RozenbergMichael Rozenberg
98.6k1590189
edited Jan 4 at 21:17
answered Dec 31 '18 at 11:43
Michael RozenbergMichael Rozenberg
98.6k1590189
answered Dec 31 '18 at 11:43
Michael RozenbergMichael Rozenberg
98.6k1590189
answered Dec 31 '18 at 11:43
Michael RozenbergMichael Rozenberg
98.6k1590189
98.6k1590189
$begingroup$
Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
$endgroup$
– Sandel
Dec 31 '18 at 17:49
$begingroup$
Happy New Year!
$endgroup$
– Michael Rozenberg
Dec 31 '18 at 17:50
add a comment |
$begingroup$
Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
$endgroup$
– Sandel
Dec 31 '18 at 17:49
$begingroup$
Happy New Year!
$endgroup$
– Michael Rozenberg
Dec 31 '18 at 17:50
$begingroup$
Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
$endgroup$
– Sandel
Dec 31 '18 at 17:49
$begingroup$
Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
$endgroup$
– Sandel
Dec 31 '18 at 17:49
$begingroup$
Happy New Year!
$endgroup$
– Michael Rozenberg
Dec 31 '18 at 17:50
$begingroup$
Happy New Year!
$endgroup$
– Michael Rozenberg
Dec 31 '18 at 17:50
add a comment |
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$begingroup$
And what about $n$? Is $$ngeq 1$$?
$endgroup$
– Dr. Sonnhard Graubner
Dec 31 '18 at 10:51
$begingroup$
Could you show your work, having proceeded as you mention?
$endgroup$
– amWhy
Dec 31 '18 at 21:52
$begingroup$
This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc.
$endgroup$
– Carl Mummert
Jan 11 at 13:55