prove the inequality

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prove the inequality

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#1 h Sep 7, 2010, 12:09 AM • 2 Y

Y by Adventure10, Mango247

let a, b, c are the length of the sides of a triangle ABC with $A, B, C < 90^0$ , prove that
$\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2} \ge 9R^2$

Nguyễn Sơn Hà
Teaching Mathematics for High School for Gifted Students, Hanoi National University of Education Vietnam

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The math is cool

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The math is cool

#2 h Sep 7, 2010, 4:40 AM • 1 Y

Y by Adventure10

Making $a=2R\sin A$ , $b=2R\sin B$ and $c=2R\sin C$ , the inequality is equivalent to
$\frac{4\sin^2A\sin^2B}{\sin^2C}+\frac{4\sin^2B\sin^2C}{\sin^2A}+\frac{4\sin^2C\sin^2A}{\sin^2B}\geq 9 \Leftrightarrow$
$\Leftrightarrow \sum\frac{(1-\cos^2A)(1-\cos^2B)}{1-\cos^2C}\geq \frac{9}{4}$ , with $A+B+C=\pi$
With the well-known identity $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C =1$ , and making the substitution
$X=\frac{\cos B\cos C}{\cos A}$ , $Y=\frac{\cos A\cos C}{\cos B}$ and $Z=\frac{\cos A\cos B}{\cos C}$ , we have $\cos^2A=YZ$ , $\cos^2B=XZ$ and $\cos^2C=XY$ and the inequality is equivalent to
$\sum\frac{(1-YZ)(1-XZ)}{1-XY}\geq \frac{9}{4}$ , with $XY+YZ+ZX+2XYZ=1$
Again, making the substitution $X=\frac{x}{y+z}$ , $Y=\frac{y}{x+z}$ and $Z=\frac{z}{x+y}$ , the inequality is equivalent to
$\sum\frac{(1-(\frac{y}{x+z})(\frac{z}{x+y}))(1-(\frac{x}{y+z})(\frac{z}{x+y}))}{1-(\frac{x}{y+z})(\frac{y}{x+z})}\geq \frac{9}{4} \Leftrightarrow$
$\Leftrightarrow \sum\frac{(x(x+y+z))(y(x+y+z))}{(x+y)^2(z(x+y+z))}\geq\frac{9}{4}\Leftrightarrow \sum\frac{xy(x+y+z)}{z(x+y)^2}\geq \frac{9}{4}$
But I don't know how to solve this last inequality!

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#3 h Sep 7, 2010, 6:01 AM • 2 Y

Y by Adventure10, Mango247

The math is cool wrote:

$\Leftrightarrow \sum\frac{(x(x+y+z))(y(x+y+z))}{(x+y)^2(z(x+y+z))}\geq\frac{9}{4}\Leftrightarrow \sum\frac{xy(x+y+z)}{z(x+y)^2}\geq \frac{9}{4}$
But I don't know how to solve this last inequality!

Proof of the last inequality:
This ineq is equivalent to:
$\sum\frac{1}{(ab+ac)^2} \ge \frac{9}{4abc(a+b+c)}$
This's Iran 96 inequality with $x=ab,y=bc,z=ca$

This post has been edited 1 time. Last edited by minhhoang, Sep 7, 2010, 9:27 AM

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kuing

#4 h Sep 7, 2010, 8:23 AM • 2 Y

Y by Adventure10, Mango247

minhhoang wrote:

The math is cool wrote:

$\Leftrightarrow \sum\frac{(x(x+y+z))(y(x+y+z))}{(x+y)^2(z(x+y+z))}\geq\frac{9}{4}\Leftrightarrow \sum\frac{xy(x+y+z)}{z(x+y)^2}\geq \frac{9}{4}$
But I don't know how to solve this last inequality!

Proof of the last inequality:
This ineq is equivalent to:
$\sum\frac{1}{(ab+ac)^2} \ge \frac{9}{4abc(ab+bc+ca)}$
This's Iran 96 inequality with $x=ab,y=bc,z=ca$

should be $\sum\frac{1}{(ab+ac)^2} \ge \frac{9}{4abc(a+b+c)}$ with $x=ab,y=bc,z=ca$

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kuing

#5 h Sep 7, 2010, 9:06 AM • 2 Y

Y by Adventure10, Mango247

$\sum {\frac{{a^2 b^2 }}{{c^2 }}} \ge 9R^2$

$\Leftrightarrow \sum {\frac{{\sin ^2 A\sin ^2 B}}{{\sin ^2 C}}} \ge \frac{9}{4}$

( $A \leftrightarrow \frac{{\pi - D}}{2},B \leftrightarrow \frac{{\pi - E}}{2},C \leftrightarrow \frac{{\pi - F}}{2}$ ,then DEF be a triangle)

$\Leftrightarrow \sum {\frac{{\sin ^2 \frac{{\pi - D}}{2}\sin ^2 \frac{{\pi - E}}{2}}}{{\sin ^2 \frac{{\pi - F}}{2}}}} \ge \frac{9}{4}$

$\Leftrightarrow \sum {\frac{{\left( {\frac{{1 + \cos D}}{2}} \right)\left( {\frac{{1 + \cos E}}{2}} \right)}}{{\frac{{1 + \cos F}}{2}}}} \ge \frac{9}{4}$

(let $\cos D = 1 - \frac{{2yz}}{{\left( {z + x} \right)\left( {x + y} \right)}}, \cdots \cdots,x,y,z > 0$ $\Rightarrow \frac{{1 + \cos D}}{2} = \frac{{x\left( {x + y + z} \right)}}{{\left( {z + x} \right)\left( {x + y} \right)}}, \cdots \cdots$ )

$\Leftrightarrow \sum {\frac{{xy\left( {x + y + z} \right)}}{{z\left( {x + y} \right)^2 }}} \ge \frac{9}{4}$

$\Leftrightarrow xyz\left( {x + y + z} \right)\sum {\frac{1}{{\left( {zx + yz} \right)^2 }}} \ge \frac{9}{4}$

$\Leftrightarrow {\rm{Iran 96}}$

another transform, but result same to 2#, so interesting...

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arqady

#6 h Sep 9, 2010, 3:10 AM • 2 Y

Y by Adventure10, Mango247

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=119891

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