Geometric Inequality

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41787 posts

sqing

#1 h Sep 10, 2015, 11:24 AM • 1 Y

Y by Adventure10

In $\triangle ABC$ , prove that $\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\geq \frac{2}{\sqrt{3}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).$

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arqady

#2 h Sep 10, 2015, 12:33 PM • 2 Y

Y by Adventure10, Mango247

After using Ravi's substitution we need to prove that
$3(x+y+z)(x+y)^2(x+z)^2(y+z)^2\geq4\left(\sum_{cyc}(x^2+3xy)\right)^2xyz$ .
Let $x+y+z=3u$ , $xy+xz+yz=3v^2$ and $xyz=w^3$ . Hence, we need to prove that $f(w^3)\geq0$ ,
where $f(w^3)=u(9uv^2-w^3)^2-4(3u^2+v^2)^2w^3$ .
We see that $f'(w^3)=-2u(9uv^2-w^3)-4(3u^2+v^2)^2<0$ .
Thus, $f$ gets a minimal value for a maximal value of $w^3$ , which happens for equality case of two variables.
Let $y=z=1$ . Hence, we need to prove $(x-1)^2(x+3)\geq0$ . Done!

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15849 posts

xzlbq

#3 h Sep 11, 2015, 12:01 AM • 2 Y

Y by Adventure10, Mango247

Stonger is:

$\left( {y}^{2}+10\,yz+{z}^{2} \right) \left( z+x \right) ^{2} \left( x+y \right) ^{2} \left( x+y+z \right) \geq 4\, \left( {x}^{2}+{y}^ {2}+{z}^{2}+3\,xy+3\,yz+3\,xz \right) ^{2}xyz$
BQ

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xzlbq

#4 h Sep 11, 2015, 12:20 AM • 2 Y

Y by Adventure10, Mango247

$\left( {z}^{2}+2\,xz+2\,{x}^{2}+2\,xy+{y}^{2} \right) ^{2} \left( x+y +z \right) ^{2}\geq 2/3\, \left( {z}^{2}+2\,xz+2\,{x}^{2}+{y}^{2}+2\,xy-2 \,yz \right) \left( {x}^{2}+{y}^{2}+{z}^{2}+3\,xy+3\,yz+3\,xz \right) ^{2}$
=>
In triangle,

$\frac{1}{k_a} \geq \frac{2}{3}\sqrt{3}(\frac{a}{a+b+c})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
$k_a=2bc/(b^2+c^2)m_a$

This post has been edited 1 time. Last edited by xzlbq, Sep 11, 2015, 12:22 AM

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xzlbq

#5 h Sep 11, 2015, 2:58 AM • 2 Y

Y by Adventure10, Mango247

$\frac{1}{w_a}\geq \frac{2\sqrt{3}}{3}\frac{(2h_c+2r_c+h_a+h_b)(h_c+h_a+2h_b+2r_b)(h_a+r_a)a}{\sum{((2h_c+2r_c+h_a+h_b)(h_c+h_a+2h_b+2r_b)(h_a+r_a)a)}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
deg=10

BQ

This post has been edited 3 times. Last edited by xzlbq, Sep 11, 2015, 3:09 AM

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#6 h Sep 11, 2015, 4:44 AM • 1 Y

Y by Adventure10

xzlbq wrote:

Stonger is:

$\left( {y}^{2}+10\,yz+{z}^{2} \right) \left( z+x \right) ^{2} \left( x+y \right) ^{2} \left( x+y+z \right) \geq 4\, \left( {x}^{2}+{y}^ {2}+{z}^{2}+3\,xy+3\,yz+3\,xz \right) ^{2}xyz$
BQ

We have
$(y^2+10yz+z^2)(z+x)^2(x+y)^2(x+y+z)-4(x^2+3xy+3xz+y^2+3yz+z^2)^2xyz=$
$(x+y+z)(x^2y+x^2z+xy^2-2xyz+xz^2-y^2z-yz^2)^2+4yz(2z+2y+x)(x^2-yz)^2\ge{0}$

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2032 posts

#7 h Sep 11, 2015, 5:15 AM • 1 Y

Y by Adventure10

xzlbq wrote:

$\left( {z}^{2}+2\,xz+2\,{x}^{2}+2\,xy+{y}^{2} \right) ^{2} \left( x+y +z \right) ^{2}\geq 2/3\, \left( {z}^{2}+2\,xz+2\,{x}^{2}+{y}^{2}+2\,xy-2 \,yz \right) \left( {x}^{2}+{y}^{2}+{z}^{2}+3\,xy+3\,yz+3\,xz \right) ^{2}$
=>
In triangle,

$\frac{1}{k_a} \geq \frac{2}{3}\sqrt{3}(\frac{a}{a+b+c})(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
$k_a=2bc/(b^2+c^2)m_a$

We have
$(z^2+2xz+2x^2+2xy+y^2)^2(x+y+z)^2-\frac{2}{3}(z^2+2xz+2x^2+y^2+2xy-2yz)(\sum{(x^2+3yz)})^2$
$=\frac{1}{3}(x^2+2yz)(y-z)^4+\frac{1}{3}(x^2z+y^3-2y^2z)^2+\frac{1}{3}(x^2y-2yz^2+z^3)^2+\frac{2}{3}x(y+z)(x^2+y^2-3yz+z^2)^2+\frac{1}{3}(8x^2+18xy+18xz+13y^2+20yz+13z^2)(x^2-yz)^2\ge{0}$

This post has been edited 1 time. Last edited by szl6208, Sep 11, 2015, 5:17 AM

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xzlbq

#8 h Sep 11, 2015, 5:25 AM • 2 Y

Y by Adventure10, Mango247

Stronger is:

$\frac{1}{\sqrt{h_a}} \geq \frac{\sqrt{2}}{2}(\frac{h_b+h_c}{h_a+h_b+h_c})\sqrt{\sqrt{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}$

Click to reveal hidden text

=>
$\left( {\frac {1}{\sqrt {{\it h_a}}}}+{\frac {1}{\sqrt {{\it h_b}}}}+{ \frac {1}{\sqrt {{\it h_c}}}} \right) ^{2}\geq 2\,\sqrt {3} \left( {a}^{-1} +{b}^{-1}+{c}^{-1} \right)$

BQ

This post has been edited 4 times. Last edited by xzlbq, Sep 11, 2015, 5:29 AM

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2032 posts

#9 h Sep 11, 2015, 6:19 AM • 1 Y

Y by Adventure10

xzlbq wrote:

Stronger is:

$\frac{1}{\sqrt{h_a}} \geq \frac{\sqrt{2}}{2}(\frac{h_b+h_c}{h_a+h_b+h_c})\sqrt{\sqrt{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}$

Click to reveal hidden text

=>
$\left( {\frac {1}{\sqrt {{\it h_a}}}}+{\frac {1}{\sqrt {{\it h_b}}}}+{ \frac {1}{\sqrt {{\it h_c}}}} \right) ^{2}\geq 2\,\sqrt {3} \left( {a}^{-1} +{b}^{-1}+{c}^{-1} \right)$

BQ

We have
$(z+x)^2(x+y)^2(x^2+y^2+z^2+3xy+3yz+3xz)^2-3(x+y+z)xyz(z+2x+y)^4=$
$3x^3(4x^3+y^3+z^3)(y-z)^2+yz(60x^2+7y^2+10yz+7z^2)(x^2-yz)^2+$
$\frac{1}{2}(x^2+z^2)(x^3-x^2z+y^3-y^2z)^2+\frac{1}{2}(x^2+y^2)(x^3-x^2y-yz^2+z^3)^2+$
$\frac{1}{2}z(18x^5+4x^4y+22x^4z+68x^3z^2+12x^2y^3+64x^2yz^2+44xy^2z^2+12xyz^3+10xz^4+z^5)(x-y)^2+$
$\frac{1}{2}y(18x^5+22x^4y+4x^4z+68x^3y^2+64x^2y^2z+12x^2z^3+10xy^4+12xy^3z+44xy^2z^2+y^5)(-z+x)^2+$
$\frac{47}{2}x^2z^2(xz-y^2)^2+\frac{47}{2}x^2y^2(xy-z^2)^2\ge{0}$

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15849 posts

xzlbq

#10 h Sep 11, 2015, 6:23 AM • 1 Y

Y by Adventure10

Szl,Very good!
and kill 12>=deg>=8
BQ

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arqady

#11 h Sep 14, 2015, 7:12 AM • 1 Y

Y by Adventure10

The following inequality is also true.
For all triangle prove that:
$\frac{1}{l_a}+\frac{1}{l_b}+\frac{1}{l_c}\geq \frac{2}{\sqrt{3}}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

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15849 posts

xzlbq

#12 h Sep 14, 2015, 7:34 AM • 2 Y

Y by arqady, Adventure10

arqady wrote:

The following inequality is also true.
For all triangle prove that:
$\frac{1}{l_a}+\frac{1}{l_b}+\frac{1}{l_c}\geq \frac{2}{\sqrt{3}}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

5#,give a sum,get it.

BQ

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1158 posts

#13 h Sep 16, 2015, 5:14 PM • 2 Y

Y by Adventure10, Mango247

we have :
$\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{ca}}=\frac{1}{r}$ , then
$\frac{1}{r}\geq \frac{2}{\sqrt{3}}\sum \frac{1}{a}$ (Petrovic's inequality )

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1093 posts

#14 h Sep 22, 2015, 11:48 AM • 1 Y

Y by Adventure10

$LHS=\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{a}{ah_a}+\frac{b}{bh_b}+\frac{c}{ch_c}=\frac{a+b+c}{2A}=\frac{2s}{2sr}=\frac{1}{r}$ ;where A are triangle area
$RHS=\frac{2}{\sqrt{3}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{2}{\sqrt{3}}(\frac{h_a}{ah_a}+\frac{h_b}{bh_b}+\frac{c}{ch_c})=\frac{2}{\sqrt{3}}\frac{h_a+h_b+h_c}{2A}=\frac{h_a+h_b+h_c}{sr\sqrt{3}}$
Ineq returns to:
$h_a+h_b+h_c\le s\sqrt{3}$ (1)
Because by RAVI's Substitutions result: $h_a=\frac{2\sqrt{xyz(x+y+z)}}{y+z},h_b=\frac{2\sqrt{xyz(x+y+z)}}{z+x},h_c=\frac{2\sqrt{xyz(x+y+z)}}{x+y}$ where $x,y,z>0$
It remains to show that:
$\frac{2}{y+z}+\frac{2}{z+x}+\frac{2}{x+y}\le\sqrt{3(\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xy})}$
This result using GM ineq and Cauchy-Schwarz:
$y+z\ge 2\sqrt{yz},z+x\ge 2\sqrt{zx},x+y\ge 2\sqrt{xy}\Rightarrow LHS\le\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}+\frac{1}{\sqrt{xy}}\le(by Cauchy-Schwarz)\sqrt{3(\frac{1}{yz}+\frac{1}{zx}+\frac{1}{xy})}$
The proof is ended!
_________
Marin Sandu

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#15 h Sep 23, 2015, 1:02 AM • 2 Y

Y by Adventure10, Mango247

See here (proposed problem) PP6 $:$ the bilateral inequality $\boxed{\ \frac {\sqrt 3}{R}\ \le\ \frac 1a+\frac 1b+\frac 1c\ \le\ \frac {\sqrt 3}{2r}\ }$

what is equivalently with $\boxed{\frac {r\sqrt 3}R\le \frac {h_a+h_b+h_c}{a+b+c}\le \frac {\sqrt 3}2}$ .

This post has been edited 7 times. Last edited by Virgil Nicula, Sep 23, 2015, 2:29 AM

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sqing

#16 h Sep 23, 2015, 4:41 AM • 1 Y

Y by Adventure10

sqing wrote:

In $\triangle ABC$ , prove that $\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\geq \frac{2}{\sqrt{3}}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}).$

$\sum_{cyc}\frac{1}{h_a}=\frac{1}{abc}\sum_{cyc}\frac{a^2}{sinA}\geq\frac{(\sum_{cyc}a)^2}{abc\sum_{cyc}sinA}\geq\frac{3\sum_{cyc}bc}{abc\sum_{cyc}sinA}\geq\frac{2}{\sqrt{3}}\sum_{cyc}\frac{1}{a}.$

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