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Algebra

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9 Jan 2009 11:15:20 IST

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Some Inequalities

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For x,y,z>0, prove that1. 2.

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pink_ele

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9 Jan 2009 11:29:41 IST

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we know
(a-b)^2+(b-c)^2+(a-c)^2>=0
let
a=x/y
b=y/z
c=z/x
from dis we get
2{(x/y)^2+(y/z)^2+(z/x)^2-x/z-y/x-z/y)>=0
dis gives
x/y)^2+(y/z)^2+(z/x)^2>=x/z+y/x+z/y
proved

pink_ele

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9 Jan 2009 11:31:05 IST

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same way if we take
a=y/x
b=z/y
and
c=x/z
we get d second proof

Hari Shankar

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9 Jan 2009 11:57:22 IST

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No, the second one is not as evident as the first. Please check

Conjurer

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9 Jan 2009 12:14:43 IST

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( x/y -1)^2 + ( y/z -1 )^2 + ( z/x -1 )^2 >=0

=> (x/y)^2 + (y/z)^2 + (z/x)^2 >= 2(x/y + y/z + z/x) - 3

If we prove x/y + y/z + z/x >=3 we are done and it follows from AM>=GM.

Hari Shankar

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9 Jan 2009 13:49:29 IST

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nice one. any more

Rahul Duggal

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9 Jan 2009 23:17:09 IST

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in the changed inequality in the right hand side. instead for z/x it is z/y

PLS CORRECT IF WRONG

Hari Shankar

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10 Jan 2009 09:13:44 IST

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Ok, I think i will post my solution to this. Actually, the first time i saw this on mathlinks, i glibly gave the same proof as pink_ele and by the time i realized my mistake it was too late (in other words, more than 24 hours later )

But let us assume the contrary i.e. $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} > \frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2}$

Then, by Cauchy Schwarz,

$3 \left(\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \right) > 3 \left( \frac{x^2}{y^2} + \frac{y^2}{z^2} + \frac{z^2}{x^2} \right) > \left(\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \right)^2 \\ \\<br/>\Rightarrow \left(\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \right)<3$

which contradicts AM-GM inequality that $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \ge 3$

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