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For a>0.b>0,c>0, then 1/a+1/b+1/c>9a,b,c real a+b+c=1

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rouche | Student, Undergraduate

Posted December 28, 2011 at 1:51 AM via web

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For a>0.b>0,c>0, then 1/a+1/b+1/c>9

a,b,c real

a+b+c=1

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted December 28, 2011 at 2:17 AM (Answer #1)

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You need to divide the relation a+b+c = 1 by a such that:

`1 + b/a + c/a = 1/a`

You need to divide the relation a+b+c = 1 by b such that:

`a/b + 1 + c/b = 1/b`

You need to divide the relation a+b+c = 1 by c such that:

`a/c + b/c + 1 = 1/c`

`` Adding these three relations yields:

`1/a + 1/b +1/c = 1 + b/a + c/a + a/b + 1 + c/b + a/c + b/c + 1 = `

`=3 + (ac + bc + ab + cb + ac + ba)/(abc)= `

`= 3 + 2(ab + ac + bc)/(abc) gt 3 + 2*3 = 9`

Hence, the last inequality proves that 1/a + 1/b +1/c > 9 if a+b+c=1.

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