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give me the total solution

Note by Superman Son
4 years, 5 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 - 4 years, 5 months ago

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thanks

Superman Son - 4 years, 4 months ago

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could you type these up? no offense, but i can't read any of the exponents

Matthew Lipman - 4 years, 5 months ago

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exponent is 8

Superman Son - 4 years, 4 months ago

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use AM-GM inequality

Ethan Tan - 4 years, 5 months ago

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please explain totally

Superman Son - 4 years, 5 months ago

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sorry lah... ok I'll try shortly

Ethan Tan - 4 years, 5 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 - 4 years, 4 months ago

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@Arushit Mudgal i know that

Superman Son - 4 years, 4 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 - 4 years, 4 months ago

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I think the second one is Titu's Lemma...

Brian Reinhart - 4 years, 5 months ago

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1st one is obvious by weighted AM-GM. Second one is obvious by Jensen's inequality.

Gabriel Wong - 4 years, 5 months ago

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Please type this up. I can't read it clearly.

Zawad Abdullah - 4 years, 5 months ago

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explain clearly

Mannette Kier Macalaguing - 4 years, 5 months ago

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Please don't, it is best for people to figure it out for themselves! By giving away more we learn less.

Brock West - 4 years, 4 months ago

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I thought we weren't supposed to put boring homework on this forum...

Justin Wong - 4 years, 5 months ago

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it is not my homework guys

Superman Son - 4 years, 4 months ago

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its an example from a book called excursion in mathemay=tics

Superman Son - 4 years, 4 months ago

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@Superman Son Ya! It is a famous book in India and a must for the basics.

Shourya Pandey - 4 years, 4 months ago

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I am sure you did not see the problems properly.

Shourya Pandey - 4 years, 4 months ago

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I thought we were't supposed to bring homework on this forum. boring or otherwise.

Francis Gerard Magtibay - 4 years, 5 months ago

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