give me the total solution

Note by Superman Son
6 years, 4 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, abca \geq b \geq c. Then, by Rearrangement inequality on the three sets a2,b2,c2a^{2},b^{2},c^{2} , a3,b3,c3a^{3},b^{3},c^{3} and a3,b3,c3a^{3},b^{3},c^{3}, we get

a2a3a3+b2b3b3+c2c3c3a2b3c3+a3b2c3+a3b3c2a^{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 a8+b8+c8a3b3c3(1a+1b+1c)a^{8}+b^{8}+c^{8} \geq a^{3}b^{3}c^{3} (\frac {1}{a} + \frac {1}{b} + \frac {1}{c}) And finally

a8+b8+c8a3b3c3(1a+1b+1c)\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:

a8+b8+c83>(a+b+c3)8\frac {a^{8}+b^{8}+c^{8}}{3} > (\frac {a+b+c}{3})^{8} is the same as

(a8+b8+c83)18>a+b+c3(\frac {a^{8}+b^{8}+c^{8}}{3})^{\frac {1}{8}} > \frac {a+b+c}{3} and this nothing but the generalized mean with M8(a,b,c)>M1(a,b,c)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=ca=b=c

Shourya Pandey - 6 years, 4 months ago

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thanks

superman son - 6 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 - 6 years, 4 months ago

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exponent is 8

superman son - 6 years, 4 months ago

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use AM-GM inequality

Ethan Tan - 6 years, 4 months ago

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please explain totally

superman son - 6 years, 4 months ago

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sorry lah... ok I'll try shortly

Ethan Tan - 6 years, 4 months ago

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@Ethan Tan AM GM means Arithmetic mean and geometric mean...and there is an proved inequality for that...

A Former Brilliant Member - 6 years, 4 months ago

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@A Former Brilliant Member i know that

superman son - 6 years, 4 months ago

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

Brian Reinhart - 6 years, 4 months ago

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i. By AM-GM,

28a8+38b8+38c8a2b3c328b8+38c8+38a8b2c3a328c8+38a8+38b8c2a3b3\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 a3b3c3a^3b^3c^3 gives the desired inequality.

ii. This is a special case of the power-mean inequality on three variables.

Ang Yan Sheng - 6 years, 4 months ago

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

Zawad Abdullah - 6 years, 4 months ago

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

Gabriel Wong - 6 years, 4 months ago

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explain clearly

Mannette Kier Macalaguing - 6 years, 4 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 - 6 years, 4 months ago

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

Justin Wong - 6 years, 4 months ago

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it is not my homework guys

superman son - 6 years, 4 months ago

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

superman son - 6 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 - 6 years, 4 months ago

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I am sure you did not see the problems properly.

Shourya Pandey - 6 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 - 6 years, 4 months ago

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