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xzlbq

#1 h Nov 1, 2010, 1:27 AM • 2 Y

Y by Adventure10, Mango247

$x,y,z>0$ ,
prove that:
$\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}} \geq \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}$
BQ

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xzlbq

#2 h Nov 1, 2010, 1:40 AM • 2 Y

Y by Adventure10, Mango247

$/2\,\sqrt {6} \left( \sqrt {{\frac {{\it x1}}{{\it x1}+2}}}+\sqrt {{\frac {{\it x2}}{{\it x2}+2}}}+\sqrt {{\frac {{\it x3}}{{\it x3}+2}}}+ \sqrt {{\frac {{\it x4}}{{\it x4}+2}}} \right) \leq {\frac { \left( \sqrt {{\it x1}}+\sqrt {{\it x2}}+\sqrt {{\it x3}}+\sqrt {{\it x4}} \right) ^{2}}{\sqrt {{\it x1}\, \left( {\it x2}+{\it x3} \right) }+ \sqrt {{\it x2}\, \left( {\it x3}+{\it x4} \right) }+\sqrt {{\it x3}\, \left( {\it x4}+{\it x1} \right) }+\sqrt {{\it x4}\, \left( {\it x1}+ {\it x2} \right) }}}$

and n-var?
BQ

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kuing

#3 h Nov 1, 2010, 5:05 AM • 2 Y

Y by Adventure10, Mango247

xzlbq wrote:

$x,y,z>0$ ,
prove that:
$\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}} \geq \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}$
BQ

stronger one

$\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}} \geq 2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}}$ $\geq \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}$

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kuing

#4 h Nov 1, 2010, 12:20 PM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

xzlbq wrote:

$x,y,z>0$ ,
prove that:
$\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}} \geq \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}$
BQ

stronger one

$\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}} \geq 2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}}$ $\geq \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}$

the right side is well-know, and I get a prove for the left side but so ugly, who can prove the left side simple?

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#5 h Nov 1, 2010, 2:17 PM • 1 Y

Y by Adventure10

kuing wrote:

stronger one

$\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}} \geq 2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}}$ $\geq \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}$

I have a classical solution for the left. I will post it later

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#6 h Nov 1, 2010, 3:42 PM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

stronger one

$\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}} \geq 2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}}$ $\geq \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}$

It is obvious from AM-GM and Cauchy.
$\Longleftrightarrow (\sqrt{x}+\sqrt{y}+\sqrt{z})^2\geq 2\sqrt{(x+y+z)(xy+yz+zx)}\cdot\sum_{cyc}{\sqrt{\frac{x}{(x+y)(x+z)}}}$
And notice that
$(xy+yz+zx)(x+y+z)\leq \frac{9}{8}(x+y)(y+z)(z+x)$
we just need to show that
$\frac{3}{\sqrt{2}}\sum_{cyc}{\sqrt{x(y+z)}}\leq (\sqrt{x}+\sqrt{y}+\sqrt{z})^2$
From Cauchy we have
$\left(\frac{3}{\sqrt{2}}\sum_{cyc}{\sqrt{x(y+z)}}\right)^2\leq \frac{9}{4}\sum_{cyc}{\sqrt{x}}\cdot\sum_{cyc}{\sqrt{x}(y+z)}$
Thus,it's enough to show that
$\sum_{cyc}{a(b^2+c^2)}\leq \frac{9}{4}(a+b+c)^3$
where $\sqrt{x}=a......$ ,and this is quite easy..

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kuing

#7 h Nov 1, 2010, 3:48 PM • 2 Y

Y by Adventure10, Mango247

let me show my prove:

$x,y,z>0$
$\frac{{(\sqrt x + \sqrt y + \sqrt z )^2 }}{{\sqrt {x(y + z)} + \sqrt {y(z + x)} + \sqrt {z(x + y)} }} \ge 2\sqrt {\frac{{(x + y + z)(xy + yz + zx)}}{{(x + y)(y + z)(z + x)}}} .$
let $x = a^2 ,y = b^2 ,z = c^2 ,a,b,c > 0$ , then the inequality can be written as
$\frac{{\left( {\sum a } \right)^4 }}{{\left( {\sum {a\sqrt {b^2 + c^2 } } } \right)^2 }} \ge \frac{{4\sum {a^2 } \sum {a^2 b^2 } }}{{\prod {\left( {a^2 + b^2 } \right)} }},$ or $\frac{{\left( {\sum a } \right)^2 }}{{\left( {\sum {\frac{a}{{a + b + c}}\sqrt {b^2 + c^2 } } } \right)^2 }} \ge \frac{{4\sum {a^2 } \sum {a^2 b^2 } }}{{\prod {\left( {a^2 + b^2 } \right)} }}.$
by Jensen's Inequality, we have
$\sum {\frac{a}{{a + b + c}}\sqrt {b^2 + c^2 } } \le \sqrt {\sum {\frac{{a\left( {b^2 + c^2 } \right)}}{{a + b + c}}} } = \sqrt {\frac{{\sum {a\left( {b^2 + c^2 } \right)} }}{{a + b + c}}} ,$ so we just need to prove that
$\begin{align*} &\frac{{\left( {\sum a } \right)^3 }}{{\sum {a\left( {b^2 + c^2 } \right)} }} \ge \frac{{4\sum {a^2 } \sum {a^2 b^2 } }}{{\prod {\left( {a^2 + b^2 } \right)} }} \\ \iff & \frac{{\left( {\sum a } \right)^3 }}{{4\sum {a\left( {b^2 + c^2 } \right)} }} \ge \frac{{\prod {\left( {a^2 + b^2 } \right)} + a^2 b^2 c^2 }}{{\prod {\left( {a^2 + b^2 } \right)} }} \\ \iff & \frac{{\left( {\sum a } \right)^3 }}{{4\sum {a\left( {b^2 + c^2 } \right)} }} - 1 \ge \frac{{a^2 b^2 c^2 }}{{\prod {\left( {a^2 + b^2 } \right)} }} \\ \iff & \frac{{\sum {a^3 } - \sum {a\left( {b^2 + c^2 } \right)} + 6abc}}{{4\sum {a\left( {b^2 + c^2 } \right)} }} \ge \frac{{a^2 b^2 c^2 }}{{\prod {\left( {a^2 + b^2 } \right)} }}. \end{align*}$
by Schur's Inequaliy, we have
$\sum {a^3} - \sum {a\left( {b^2 + c^2 } \right)} + 6abc \ge 3abc,$ so we just need to prove that
$\begin{align*} & \frac{{3abc}}{{4\sum {a\left( {b^2 + c^2 } \right)} }} \ge \frac{{a^2 b^2 c^2 }}{{\prod {\left( {a^2 + b^2 } \right)} }} \\ \iff & 3\prod {\left( {a^2 + b^2 } \right)} \ge 4abc\sum {a\left( {b^2 + c^2 } \right)} \\ \iff & \sum {c^2 (3a^2 + 2ab + 3b^2 )(a - b)^2 } \ge 0, \end{align*}$ obvious true. the prove is completed.

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kuing

#8 h Nov 1, 2010, 4:00 PM • 2 Y

Y by Adventure10, Mango247

Tourish wrote:

kuing wrote:

stronger one

$\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}} \geq 2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}}$ $\geq \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}$

It is obvious from AM-GM and Cauchy.
$\Longleftrightarrow (\sqrt{x}+\sqrt{y}+\sqrt{z})^2\geq 2\sqrt{(x+y+z)(xy+yz+zx)}\cdot\sum_{cyc}{\sqrt{\frac{x}{(x+y)(x+z)}}}$
And notice that
$(xy+yz+zx)(x+y+z)\leq \frac{9}{8}(x+y)(y+z)(z+x)$
we just need to show that
$\frac{3}{\sqrt{2}}\sum_{cyc}{\sqrt{x(y+z)}}\leq (\sqrt{x}+\sqrt{y}+\sqrt{z})^2$
From Cauchy we have
$\left(\frac{3}{\sqrt{2}}\sum_{cyc}{\sqrt{x(y+z)}}\right)^2\leq \frac{9}{4}\sum_{cyc}{\sqrt{x}}\cdot\sum_{cyc}{\sqrt{x}(y+z)}$
Thus,it's enough to show that
$\sum_{cyc}{a(b^2+c^2)}\leq \frac{9}{4}(a+b+c)^3$
where $\sqrt{x}=a......$ ,and this is quite easy..

so sorry, this ineq $\frac{3}{\sqrt{2}}\sum_{cyc}{\sqrt{x(y+z)}}\leq (\sqrt{x}+\sqrt{y}+\sqrt{z})^2$ does not hold.

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#9 h Nov 1, 2010, 4:18 PM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

so sorry, this ineq $\frac{3}{\sqrt{2}}\sum_{cyc}{\sqrt{x(y+z)}}\leq (\sqrt{x}+\sqrt{y}+\sqrt{z})^2$ does not hold.

Oops,so sorry for my stupid mistake...I just calculate that $(\frac{3}{\sqrt{2}})^2=\frac{9}{4}$ ...

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#10 h Nov 1, 2010, 4:22 PM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

kuing wrote:

stronger one

$\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}} \geq 2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}}$ $\geq \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}$

The right hand side:
Use Cauchy-Schwarz ineq, we have:
$\sqrt {\frac{x}{{x + y}}} + \sqrt {\frac{y}{{y + z}}} + \sqrt {\frac{z}{{z + x}}}$
$\le \sqrt {(x + z + y + x + z + y)(\frac{x}{{(x + y)(x + z)}} + \frac{y}{{(y + z)(y + x)}} + \frac{z}{{(z + x)(z + y)}})}$
$= 2\sqrt {\frac{{(x + y + z)(xy + yz + zx)}}{{(x + y)(y + z)(z + x)}}}$

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Kimpul

#11 h Nov 1, 2010, 11:40 PM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

let me show my prove:

$x,y,z>0$
$\frac{{(\sqrt x + \sqrt y + \sqrt z )^2 }}{{\sqrt {x(y + z)} + \sqrt {y(z + x)} + \sqrt {z(x + y)} }} \ge 2\sqrt {\frac{{(x + y + z)(xy + yz + zx)}}{{(x + y)(y + z)(z + x)}}} .$
let $x = a^2 ,y = b^2 ,z = c^2 ,a,b,c > 0$ , then the inequality can be written as
$\frac{{\left( {\sum a } \right)^4 }}{{\left( {\sum {a\sqrt {b^2 + c^2 } } } \right)^2 }} \ge \frac{{4\sum {a^2 } \sum {a^2 b^2 } }}{{\prod {\left( {a^2 + b^2 } \right)} }},$ or $\frac{{\left( {\sum a } \right)^2 }}{{\left( {\sum {\frac{a}{{a + b + c}}\sqrt {b^2 + c^2 } } } \right)^2 }} \ge \frac{{4\sum {a^2 } \sum {a^2 b^2 } }}{{\prod {\left( {a^2 + b^2 } \right)} }}.$
by Jensen's Inequality, we have
$\sum {\frac{a}{{a + b + c}}\sqrt {b^2 + c^2 } } \le \sqrt {\sum {\frac{{a\left( {b^2 + c^2 } \right)}}{{a + b + c}}} } = \sqrt {\frac{{\sum {a\left( {b^2 + c^2 } \right)} }}{{a + b + c}}} ,$ so we just need to prove that
$\begin{align*} &\frac{{\left( {\sum a } \right)^3 }}{{\sum {a\left( {b^2 + c^2 } \right)} }} \ge \frac{{4\sum {a^2 } \sum {a^2 b^2 } }}{{\prod {\left( {a^2 + b^2 } \right)} }} \\ \iff & \frac{{\left( {\sum a } \right)^3 }}{{4\sum {a\left( {b^2 + c^2 } \right)} }} \ge \frac{{\prod {\left( {a^2 + b^2 } \right)} + a^2 b^2 c^2 }}{{\prod {\left( {a^2 + b^2 } \right)} }} \\ \iff & \frac{{\left( {\sum a } \right)^3 }}{{4\sum {a\left( {b^2 + c^2 } \right)} }} - 1 \ge \frac{{a^2 b^2 c^2 }}{{\prod {\left( {a^2 + b^2 } \right)} }} \\ \iff & \frac{{\sum {a^3 } - \sum {a\left( {b^2 + c^2 } \right)} + 6abc}}{{4\sum {a\left( {b^2 + c^2 } \right)} }} \ge \frac{{a^2 b^2 c^2 }}{{\prod {\left( {a^2 + b^2 } \right)} }}. \end{align*}$
by Schur's Inequaliy, we have
$\sum {a^3} - \sum {a\left( {b^2 + c^2 } \right)} + 6abc \ge 3abc,$ so we just need to prove that
$\begin{align*} & \frac{{3abc}}{{4\sum {a\left( {b^2 + c^2 } \right)} }} \ge \frac{{a^2 b^2 c^2 }}{{\prod {\left( {a^2 + b^2 } \right)} }} \\ \iff & 3\prod {\left( {a^2 + b^2 } \right)} \ge 4abc\sum {a\left( {b^2 + c^2 } \right)} \\ \iff & \sum {c^2 (3a^2 + 2ab + 3b^2 )(a - b)^2 } \ge 0, \end{align*}$ obvious true. the prove is completed.

I think your proof is great, not ugly at all.

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#12 h Nov 2, 2010, 2:51 AM • 1 Y

Y by Adventure10

kuing wrote:

stronger one

$\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}} \geq 2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}}$ $\geq \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}$

This is my proof to the left hand side inequality.

Write the inequality as
$4\left( \sum x \right) \left( \sum xy\right) \left[\sum \sqrt{\frac{x}{(x+y)(x+z)}}\right]^2 \le \left( \sum \sqrt{x}\right)^4.$
Using Cauchy-Schwarz inequality, we have
$\left[\sum \sqrt{\frac{x}{(x+y)(x+z)}}\right]^2 \le \left( \sum \sqrt{x}\right) \left[ \sum \frac{\sqrt{x}}{(x+y)(x+z)}\right].$
Therefore, it suffices to prove that
$4\left( \sum x\right) \left( \sum xy\right) \left[ \sum \frac{\sqrt{x}}{(x+y)(x+z)}\right] \le \left( \sum \sqrt{x}\right)^3.$
Now, using AM-GM inequality, one has
$\begin{aligned} \frac{4(x+y+z)(xy+yz+zx)}{(x+y)(x+z)}&=4(y+z)+\frac{4xyz}{(x+y)(x+z)} \\ &\le 4(y+z)+\frac{4xyz}{2\sqrt{xy}\cdot 2\sqrt{xz}} \\ &=4(y+z)+\sqrt{yz}.\end{aligned}$
It follows that
$\sum \frac{4\sqrt{x}\left( \sum x\right)\left( \sum xy \right)}{(x+y)(x+z)} \le \sum \left[ 4\sqrt{x}(y+z)+ \sqrt{xyz}\right],$
and we come up with the inequality
$4\sum \sqrt{x}(y+z) +3\sqrt{xyz} \le \left( \sum \sqrt{x}\right)^3,$
which is Schur's inequality applied for $\sqrt{x},$ $\sqrt{y},$ $\sqrt{z}.$

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kuing

#13 h Nov 2, 2010, 3:19 AM • 2 Y

Y by Adventure10, Mango247

can_hang2007 wrote:

kuing wrote:

stronger one

$\frac{(\sqrt{x}+\sqrt{y}+\sqrt{z})^2}{\sqrt{x(y+z)}+\sqrt{y(z+x)}+\sqrt{z(x+y)}} \geq 2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}}$ $\geq \sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}$

This is my proof to the left hand side inequality.

Write the inequality as
$4\left( \sum x \right) \left( \sum xy\right) \left[\sum \sqrt{\frac{x}{(x+y)(x+z)}}\right]^2 \le \left( \sum \sqrt{x}\right)^4.$
Using Cauchy-Schwarz inequality, we have
$\left[\sum \sqrt{\frac{x}{(x+y)(x+z)}}\right]^2 \le \left( \sum \sqrt{x}\right) \left[ \sum \frac{\sqrt{x}}{(x+y)(x+z)}\right].$
Therefore, it suffices to prove that
$4\left( \sum x\right) \left( \sum xy\right) \left[ \sum \frac{\sqrt{x}}{(x+y)(x+z)}\right] \le \left( \sum \sqrt{x}\right)^3.$
Now, using AM-GM inequality, one has
$\begin{aligned} \frac{4(x+y+z)(xy+yz+zx)}{(x+y)(x+z)}&=4(y+z)+\frac{4xyz}{(x+y)(x+z)} \\ &\le 4(y+z)+\frac{4xyz}{2\sqrt{xy}\cdot 2\sqrt{xz}} \\ &=4(y+z)+\sqrt{yz}.\end{aligned}$
It follows that
$\sum \frac{4\sqrt{x}\left( \sum x\right)\left( \sum xy \right)}{(x+y)(x+z)} \le \sum \left[ 4\sqrt{x}(y+z)+ \sqrt{xyz}\right],$
and we come up with the inequality
$4\sum \sqrt{x}(y+z) +3\sqrt{xyz} \le \left( \sum \sqrt{x}\right)^3,$
which is Schur's inequality applied for $\sqrt{x},$ $\sqrt{y},$ $\sqrt{z}.$

great prove! thank you, can

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kuing

#14 h Nov 2, 2010, 3:44 AM • 1 Y

Y by Adventure10

Kimpul wrote:

I think your proof is great, not ugly at all.

the ugly because of the last inequality I expand it then get the sos form, that's some calculation in it.

and can's prove is great indeed, because just CS, A-G and Schur, not so much calculation ...

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Kimpul

#15 h Nov 2, 2010, 4:55 AM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

Kimpul wrote:

I think your proof is great, not ugly at all.

the ugly because of the last inequality I expand it then get the sos form, that's some calculation in it.

and can's prove is great indeed, because just CS, A-G and Schur, not so much calculation ...

Oh, I see. But I still think the use of jensen and schur is great, I did not see it,
and yes, can's solution is better. I have to agree on that one.

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kuing

#16 h Nov 2, 2010, 5:10 AM • 2 Y

Y by Adventure10, Mango247

in fact, in my prove, the "jensen inequality's step" can be written as CS

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