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give me the total solution

Note by Superman Son
3 years, 7 months ago

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Diganta B., please don't forget to mention \(\geq\) instead of \(>\) in your question.

For the first one: Assume, by symmetry, \(a \geq b \geq c\). Then, by Rearrangement inequality on the three sets \(a^{2},b^{2},c^{2}\) , \(a^{3},b^{3},c^{3}\) and \(a^{3},b^{3},c^{3}\), we get

\(a^{2}a^{3}a^{3}+b^{2}b^{3}b^{3}+c^{2}c^{3}c^{3}\geq a^{2}b^{3}c^{3}+a^{3}b^{2}c^{3}+a^{3}b^{3}c^{2}\)

which gives \(a^{8}+b^{8}+c^{8} \geq a^{3}b^{3}c^{3} (\frac {1}{a} + \frac {1}{b} + \frac {1}{c})\) And finally

\(\frac {a^{8}+b^{8}+c^{8}}{a^{3}b^{3}c^{3}} \geq (\frac {1}{a} + \frac {1}{b} + \frac {1}{c})\)

For the second question:

\(\frac {a^{8}+b^{8}+c^{8}}{3} > (\frac {a+b+c}{3})^{8}\) is the same as

\((\frac {a^{8}+b^{8}+c^{8}}{3})^{\frac {1}{8}} > \frac {a+b+c}{3}\) and this nothing but the generalized mean with \(M_{8}(a,b,c) > M_{1}(a,b,c)\). I hope you know about the generalized mean inequality.

Equality in both cases holds iff \(a=b=c\) Shourya Pandey · 3 years, 7 months ago

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@Shourya Pandey thanks Superman Son · 3 years, 7 months ago

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could you type these up? no offense, but i can't read any of the exponents Matthew Lipman · 3 years, 7 months ago

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@Matthew Lipman exponent is 8 Superman Son · 3 years, 7 months ago

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use AM-GM inequality Ethan Tan · 3 years, 7 months ago

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@Ethan Tan please explain totally Superman Son · 3 years, 7 months ago

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@Superman Son sorry lah... ok I'll try shortly Ethan Tan · 3 years, 7 months ago

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@Ethan Tan AM GM means Arithmetic mean and geometric mean...and there is an proved inequality for that... Arushit Mudgal · 3 years, 7 months ago

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@Arushit Mudgal i know that Superman Son · 3 years, 7 months ago

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i. By AM-GM,

\[\frac28a^8+\frac38b^8+\frac38c^8\geq a^2b^3c^3\\ \frac28b^8+\frac38c^8+\frac38a^8\geq b^2c^3a^3\\ \frac28c^8+\frac38a^8+\frac38b^8\geq c^2a^3b^3\]

Adding these three inequalities and dividing by \(a^3b^3c^3\) gives the desired inequality.

ii. This is a special case of the power-mean inequality on three variables. Ang Yan Sheng · 3 years, 7 months ago

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I think the second one is Titu's Lemma... Brian Reinhart · 3 years, 7 months ago

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1st one is obvious by weighted AM-GM. Second one is obvious by Jensen's inequality. Gabriel Wong · 3 years, 7 months ago

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Please type this up. I can't read it clearly. Zawad Abdullah · 3 years, 7 months ago

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explain clearly Mannette Kier Macalaguing · 3 years, 7 months ago

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@Mannette Kier Macalaguing Please don't, it is best for people to figure it out for themselves! By giving away more we learn less. Brock West · 3 years, 7 months ago

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I thought we weren't supposed to put boring homework on this forum... Justin Wong · 3 years, 7 months ago

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@Justin Wong it is not my homework guys Superman Son · 3 years, 7 months ago

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@Superman Son its an example from a book called excursion in mathemay=tics Superman Son · 3 years, 7 months ago

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@Superman Son Ya! It is a famous book in India and a must for the basics. Shourya Pandey · 3 years, 7 months ago

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@Justin Wong I am sure you did not see the problems properly. Shourya Pandey · 3 years, 7 months ago

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@Justin Wong I thought we were't supposed to bring homework on this forum. boring or otherwise. Francis Gerard Magtibay · 3 years, 7 months ago

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