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






active

oldest

votes


















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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057596%2fprove-x-12-x-23-x-n-1n-frac1x-12-x-23-x-n-1n%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057596%2fprove-x-12-x-23-x-n-1n-frac1x-12-x-23-x-n-1n%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
Read more

1300-talet

Image
Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
Read more

Display a custom attribute below product name in the front-end Magento 1.9.3.8

Image
0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
Read more