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 -...












1














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







share|cite|improve this question
























  • And what about $n$? Is $$ngeq 1$$?
    – Dr. Sonnhard Graubner
    Dec 31 '18 at 10:51










  • Could you show your work, having proceeded as you mention?
    – amWhy
    Dec 31 '18 at 21:52
















1














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







share|cite|improve this question
























  • And what about $n$? Is $$ngeq 1$$?
    – Dr. Sonnhard Graubner
    Dec 31 '18 at 10:51










  • Could you show your work, having proceeded as you mention?
    – amWhy
    Dec 31 '18 at 21:52














1












1








1


1





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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 10:54







Sandel

















asked Dec 31 '18 at 10:46









SandelSandel

1785




1785












  • And what about $n$? Is $$ngeq 1$$?
    – Dr. Sonnhard Graubner
    Dec 31 '18 at 10:51










  • Could you show your work, having proceeded as you mention?
    – amWhy
    Dec 31 '18 at 21:52


















  • And what about $n$? Is $$ngeq 1$$?
    – Dr. Sonnhard Graubner
    Dec 31 '18 at 10:51










  • Could you show your work, having proceeded as you mention?
    – amWhy
    Dec 31 '18 at 21:52
















And what about $n$? Is $$ngeq 1$$?
– Dr. Sonnhard Graubner
Dec 31 '18 at 10:51




And what about $n$? Is $$ngeq 1$$?
– Dr. Sonnhard Graubner
Dec 31 '18 at 10:51












Could you show your work, having proceeded as you mention?
– amWhy
Dec 31 '18 at 21:52




Could you show your work, having proceeded as you mention?
– amWhy
Dec 31 '18 at 21:52










1 Answer
1






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1














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
$$sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2}left(prodlimits_{k=1}^{n-1}x_k^{k+1}cdotfrac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}right)^{frac{2}{n(n+1)}}=frac{n(n+1)}{2}.$$






share|cite|improve this answer























  • Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
    – Sandel
    Dec 31 '18 at 17:49










  • Happy New Year!
    – Michael Rozenberg
    Dec 31 '18 at 17:50











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1 Answer
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active

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1














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
$$sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2}left(prodlimits_{k=1}^{n-1}x_k^{k+1}cdotfrac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}right)^{frac{2}{n(n+1)}}=frac{n(n+1)}{2}.$$






share|cite|improve this answer























  • Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
    – Sandel
    Dec 31 '18 at 17:49










  • Happy New Year!
    – Michael Rozenberg
    Dec 31 '18 at 17:50
















1














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
$$sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2}left(prodlimits_{k=1}^{n-1}x_k^{k+1}cdotfrac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}right)^{frac{2}{n(n+1)}}=frac{n(n+1)}{2}.$$






share|cite|improve this answer























  • Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
    – Sandel
    Dec 31 '18 at 17:49










  • Happy New Year!
    – Michael Rozenberg
    Dec 31 '18 at 17:50














1












1








1






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
$$sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2}left(prodlimits_{k=1}^{n-1}x_k^{k+1}cdotfrac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}right)^{frac{2}{n(n+1)}}=frac{n(n+1)}{2}.$$






share|cite|improve this answer














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
$$sum_{k=1}^{n-1}(k+1)x_k+frac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}geqfrac{n(n+1)}{2}left(prodlimits_{k=1}^{n-1}x_k^{k+1}cdotfrac{1}{prodlimits_{k=1}^{n-1}x_k^{k+1}}right)^{frac{2}{n(n+1)}}=frac{n(n+1)}{2}.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 21:17

























answered Dec 31 '18 at 11:43









Michael RozenbergMichael Rozenberg

97.5k1589188




97.5k1589188












  • Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
    – Sandel
    Dec 31 '18 at 17:49










  • Happy New Year!
    – Michael Rozenberg
    Dec 31 '18 at 17:50


















  • Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
    – Sandel
    Dec 31 '18 at 17:49










  • Happy New Year!
    – Michael Rozenberg
    Dec 31 '18 at 17:50
















Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
– Sandel
Dec 31 '18 at 17:49




Thank you a lot, this was extremely nice and helpful! And I wish you a happy New Year! :)
– Sandel
Dec 31 '18 at 17:49












Happy New Year!
– Michael Rozenberg
Dec 31 '18 at 17:50




Happy New Year!
– Michael Rozenberg
Dec 31 '18 at 17:50


















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