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Source: Alberobello

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826 posts

#1 h Jan 1, 2009, 11:09 PM • 2 Y

Y by Adventure10, Mango247

Let $a, b, c, k$ be non-negative real numbers. Prove o disprove that

$\sum[\frac {(a^2 - bc)(\sqrt {2a^2 + b^2 + c^2} + k)}{2a^2 + b^2 + c^2}] \ge 0$ .

__________

:

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#2 h Dec 9, 2009, 9:36 PM • 2 Y

Y by Adventure10, Mango247

Ligouras wrote:

Let $a, b, c, k$ be non-negative real numbers. Prove o disprove that

$\sum[\frac {(a^2 - bc)(\sqrt {2a^2 + b^2 + c^2} + k)}{2a^2 + b^2 + c^2}] \ge 0$ .

__________

:

Who can solve this?

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1008 posts

kuing

#3 h Dec 10, 2009, 4:09 AM • 2 Y

Y by Adventure10, Mango247

Ligouras wrote:

Let $a, b, c, k$ be non-negative real numbers. Prove o disprove that

$\sum[\frac {(a^2 - bc)(\sqrt {2a^2 + b^2 + c^2} + k)}{2a^2 + b^2 + c^2}] \ge 0$ .

__________

:

we can prove that

$\sum{\frac{a^2-bc}{2a^2+b^2+c^2}} \ge 0$

because

$\sum{\frac{a^2-bc}{2a^2+b^2+c^2}}$

$= \frac{{\left( {\sum {\left( {a - b} \right)^2 } } \right)\left( {2\sum {a^2 \left( {a - b} \right)\left( {a - c} \right)} + 3\sum {c^2 \left( {a - b} \right)^2 } } \right)}}{{4\prod {\left( {2a^2 + b^2 + c^2 } \right)} }},$

so, we just need to prove that when $k=0$ .

to be continue...

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#4 h Dec 10, 2009, 4:36 AM • 1 Y

Y by Adventure10

kuing wrote:

so, we just need to prove that when $k = 0$ .

why?

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theSA

#5 h Dec 10, 2009, 4:59 AM • 1 Y

Y by Adventure10

kuing wrote:

Ligouras wrote:

Let $a, b, c, k$ be non-negative real numbers. Prove o disprove that

$\sum[\frac {(a^2 - bc)(\sqrt {2a^2 + b^2 + c^2} + k)}{2a^2 + b^2 + c^2}] \ge 0$ .

__________

:

we can prove that

$\sum{\frac {a^2 - bc}{2a^2 + b^2 + c^2}} \ge 0$

because

$\sum{\frac {a^2 - bc}{2a^2 + b^2 + c^2}}$
$= \frac {{\left( {\sum {\left( {a - b} \right)^2 } } \right)\left( {2\sum {a^2 \left( {a - b} \right)\left( {a - c} \right)} + 3\sum {c^2 \left( {a - b} \right)^2 } } \right)}}{{4\prod {\left( {2a^2 + b^2 + c^2 } \right)} }},$

I think this prove is enough. since all numbers are nonnegative,then the second part is always bigger then or egual to 0.
I think...

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1008 posts

kuing

#6 h Dec 10, 2009, 5:24 AM • 2 Y

Y by Adventure10, Mango247

hedeng123 wrote:

kuing wrote:

so, we just need to prove that when $k = 0$ .

why?

because $k \ge 0$ and

$\sum{\frac {(a^{2} - bc)(\sqrt {2a^{2} + b^{2} + c^{2}} + k)}{2a^{2} + b^{2} + c^{2}}}$

$= \sum{\frac {a^{2} - bc}{\sqrt {2a^{2} + b^{2} + c^{2}}}} + k\sum{\frac {a^{2} - bc}{2a^{2} + b^{2} + c^{2}}},$

I was proved $\sum{\frac {a^{2} - bc}{2a^{2} + b^{2} + c^{2}}} \ge 0$ is hold, so after we just need to prove that

$\sum{\frac {a^{2} - bc}{\sqrt {2a^{2} + b^{2} + c^{2}}}} \ge 0.$

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1008 posts

kuing

#7 h Dec 10, 2009, 5:36 AM • 2 Y

Y by Adventure10, Mango247

theSA wrote:

kuing wrote:

Ligouras wrote:

Let $a, b, c, k$ be non-negative real numbers. Prove o disprove that

$\sum[\frac {(a^2 - bc)(\sqrt {2a^2 + b^2 + c^2} + k)}{2a^2 + b^2 + c^2}] \ge 0$ .

__________

:

we can prove that

$\sum{\frac {a^2 - bc}{2a^2 + b^2 + c^2}} \ge 0$

because

$\sum{\frac {a^2 - bc}{2a^2 + b^2 + c^2}}$
$= \frac {{\left( {\sum {\left( {a - b} \right)^2 } } \right)\left( {2\sum {a^2 \left( {a - b} \right)\left( {a - c} \right)} + 3\sum {c^2 \left( {a - b} \right)^2 } } \right)}}{{4\prod {\left( {2a^2 + b^2 + c^2 } \right)} }},$

I think this prove is enough. since all numbers are nonnegative,then the second part is always bigger then or egual to 0.
I think...

I don't think so.

although we have $\sum{\frac {a^2 - bc}{2a^2 + b^2 + c^2}} \ge 0$ and $f(a,b,c) \ge 0$ , but if $f(a,b,c)$ is not a constant, there is not enough to show that $\sum{\frac {(a^2 - bc)f(a,b,c)}{2a^2 + b^2 + c^2}} \ge 0.$

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1008 posts

kuing

#8 h Dec 10, 2009, 6:12 AM • 1 Y

Y by Adventure10

continue

but I will still use $\sum{\frac {a^{2} - bc}{2a^{2} + b^{2} + c^{2}}}\ge 0$ to prove that $\sum{\frac {a^{2} - bc}{\sqrt {2a^{2} + b^{2} + c^{2}}}}\ge 0.$

WLOG, assume that $a\ge b \ge c$ , that we have

$\sqrt {2a^2 + b^2 + c^2} \ge \sqrt {a^2 + 2b^2 + c^2} \ge \sqrt {a^2 + b^2 + 2c^2},$

because of

$\frac {a^{2} - bc}{2a^{2} + b^{2} + c^{2}} - \frac {b^{2} - ca}{a^{2} + 2b^{2} + c^{2}} = \frac {(a - b)(c(a - b)^2 + \sum{a(a + b)(a + c)})}{(2a^{2} + b^{2} + c^{2})(a^{2} + 2b^{2} + c^{2})} \ge 0$

and

$\frac {b^{2} - bc}{a^{2} + 2b^{2} + c^{2}} - \frac {c^{2} - ab}{a^{2} + b^{2} + 2c^{2}} = \frac {(b - c)(a(b - c)^2 + \sum{a(a + b)(a + c)})}{(a^{2} + 2b^{2} + c^{2})(a^{2} + b^{2} + 2c^{2})} \ge 0,$

so we have

$\frac {a^{2} - bc}{2a^{2} + b^{2} + c^{2}} \ge \frac {b^{2} - ca}{a^{2} + 2b^{2} + c^{2}} \ge \frac {c^{2} - ab}{a^{2} + b^{2} + 2c^{2}}$ ,

thus, by Chebyshev Inequality, we have

$\sum{\frac {a^{2} - bc}{\sqrt {2a^{2} + b^{2} + c^{2}}}}$

$= \sum{\left(\frac {a^{2} - bc}{2a^{2} + b^{2} + c^{2}}\cdot \sqrt {2a^2 + b^2 + c^2}\right)}$

$ \ge \frac {1}{3}\cdot \left(\sum{\frac {a^{2} - bc}{2a^{2} + b^{2} + c^{2}}\right)\cdot \left(\sum{\sqrt {2a^2 + b^2 + c^2}}\right)}$

$\ge 0,$

the prove is complete.

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