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Algebra

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15 Jul 2008 14:16:53 IST

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Triangle Inequality

None

If a,b, c are sides of a triangle prove that

$\sqrt [3] {\frac{a^3+b^3+c^3 + 3abc}{2}} \ge max(a,b,c)$

Also, let me add that only honest attempts are appreciated. If you have already seen the solution somewhere wait a while for others to try it out. Thank you.

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Comments (7)

Hari Shankar

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15 Jul 2008 20:06:29 IST

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I am not allowing this prob do be drowned in spam

®µD®A

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15 Jul 2008 20:22:59 IST

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Sir i think many of them have already seen this problem . It is a very good one.

Anant Kumar

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15 Jul 2008 21:40:38 IST

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i think i shall wait for, may be, tomorrow morn.. then we can have the solution

Hari Shankar

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16 Jul 2008 09:42:57 IST

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anybody with a solution is welcome to post

Nikhil Bhattacharya

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16 Jul 2008 09:52:49 IST

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$\text{Let,any of a,b,c be maximum}$

$\text{I am taking it as }a$

$\text{To Prove}$

$\sqrt[3]{\frac{a^3+b^3+c^3+3abc}{2}}\ge a$

$a^3+b^3+c^3+3abc\ge 2a^3$

$b^3+c^3+3abc\ge a^3$

$\text{Now,I assume }b^3+c^3+3abc\ge a^3$

$(b+c)(b^2+c^2-bc)+3abc\ge a^3$

$\Rightarrow (b+c)>a$

$\text{Since, above is true}\\\\\text{our asuumption is also true}\\\\\text{hence,proved}$

Hari Shankar

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Joined: 28 Feb 2007

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16 Jul 2008 10:50:58 IST

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On similar lines, but a bit shortened:

Suppose a is the greatest of the sides. Then as above we have to prove that

b^3+c^3+3abc>a^3

Rewrite that as having to prove that

b^3+c^3-a^3-3(-a)bc>0

This easily follows from the fact that b+c>a i.e. b+c-a>0 and from a well known identitiy

Anant Kumar

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Joined: 10 Jul 2008

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16 Jul 2008 13:41:59 IST

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One of my students gave the following

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