# Interesting inequalityWatch

#1
Suppose a, b, c > 0. Prove that .
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13 years ago
#2
I think I saw it somewhere but don't remember the proof .
Is it using Cauchy inequality?
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13 years ago
#3
I find another, it is more difficult I think, but don't remember the proof
Prove:
Your inequality can be proved like that:
First prove:

Square both sides to have

Now just apply that inequality cyclic, the initial inequality is proved
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#4
Yeah, that's right. There's also a very peculiar way to prove the inequality I posted using geometry.

I'll have a go at your inequality later (hopefully).
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#5
Okay, I think I got it.

Using my inequality we can see that

Now Cauchy's inequality gives us

Where the second step holds because

Hence

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13 years ago
#6
Good move using result from your inequality
Just apply Cauchy's inequality for each pair (a+b) >= 2.sqrt(ab), (a+c) >= 2.sqrt(ac), (b+c) >= 2.sqrt(bc), for a, b, c, d >= 0
So you'll have LHS >= RHS
Equality occurs when a=b=c=d
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#7
Fine, you want a harder one then?!

(a,b,c>0 and abc=1)
Prove:
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13 years ago
#8
not now, iam so tired !
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12 years ago
#9
It's much harder than the others. But I tried few ways but noone is absolutely right . I don't really like inequality
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