Interesting inequality Watch

dvs
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#1
Report Thread starter 13 years ago
#1
Suppose a, b, c > 0. Prove that \sqrt{a^2 + ab + b^2} + \sqrt{b^2 + bc + c^2} + \sqrt{c^2 + ca + a^2} \geq \sqrt{3} (a + b + c).
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BCHL85
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#2
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I think I saw it somewhere but don't remember the proof .
Is it using Cauchy inequality?
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BCHL85
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#3
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I find another, it is more difficult I think, but don't remember the proof
Prove: \sqrt{a^4+a^2.b^2 +b^4} + \sqrt{b^4+c^2.b^2 +c^4} + \sqrt{c^4+a^2.c^2 +a^4} \geq a.\sqrt{2a^2+bc} + b.\sqrt{2b^2+ac}+ c.\sqrt{2c^2+ab}
Your inequality can be proved like that:
First prove:


\frac{\sqrt{a^2 + ab + b^2}}{3} \geq \frac{a+b}{2}
Square both sides to have


\frac{a^2 + ab + b^2}{3} \geq \frac{a^2+2ab +b^2}{4} \\

\Leftrightarrow 4(a^2 + ab +b^2} \geq 3(a^2 + 2ab + b^2) \\

\Leftrightarrow (a^2 + b^2) \geq 2ab\\

\Leftrightarrow (a-b)^2 \geq 0
Now just apply that inequality cyclic, the initial inequality is proved
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dvs
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#4
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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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dvs
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#5
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Okay, I think I got it.

Using my inequality we can see that
\sum \sqrt{a^4 + a^2 b^2 + b^4} \geq \sqrt{3} \sum a^2

Now Cauchy's inequality gives us


\begin{eqnarray}

( \sum a\sqrt{2a^2 + bc} )^2 &\leq& (\sum a^2) (\sum (2a^2 + bc)) \\

&\leq& 3 (\sum a^2)^2

\end{eqnarray}

Where the second step holds because


(\sum a^2) (\sum (2a^2 + bc)) \, \leq \, 3(\sum a^2)^2 \\

\Rightarrow ab + bc + ca \, \leq \, a^2 + b^2 + c^2 \\

\Rightarrow 0 \, \leq \, (a-b)^2 + (b-c)^2 + (c-a)^2 \\

Hence
\sum \sqrt{a^4 + a^2 b^2 + b^4} \, \geq \, \sqrt{3} \sum a^2 \, \geq \, \sum a\sqrt{2a^2 + bc}


Now how about:
\sqrt{(a+b)(c+d)} + \sqrt{(a+c)(b+d)} + \sqrt{(a+d)(b+c)} \, \geq \, 6 \sqrt[4]{abcd}
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BCHL85
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#6
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Good move using result from your inequality
Are you really serious asking about this 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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dvs
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#7
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Fine, you want a harder one then?!

(a,b,c>0 and abc=1)
Prove: \sum \frac{1}{a^3 (b + c)} \, \geq \, \frac{3}{2}
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BCHL85
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#8
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#8
not now, iam so tired !
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BCHL85
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#9
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#9
It's much harder than the others. But I tried few ways but noone is absolutely right :p:. I don't really like inequality
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