Inequality for six variables-China South East Olympiad 2009

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Inequality for six variables-China South East Olympiad 2009

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lssl

#1 h Sep 18, 2010, 10:28 AM • 1 Y

Y by Adventure10

Let $x,y,z$ be positive reals such that $\sqrt{a}=x(y-z)^2$ , $\sqrt{b}=y(z-x)^2$ and $\sqrt{c}=z(x-y)^2$ . Prove that
$a^2+b^2+c^2 \geq 2(ab+bc+ca)$

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jgnr

#2 h Sep 18, 2010, 3:22 PM • 4 Y

Y by GanzorigBattur, FERZIO, Adventure10, Mango247

Suppose $a^2+b^2+c^2<2(ab+bc+ca)$
$(a-b-c)^2<4bc$
$-2\sqrt{bc}<a-b-c<2\sqrt{bc}$
$(\sqrt{b}-\sqrt{c})^2<a<(\sqrt{b}+\sqrt{c})^2$
$|\sqrt{b}-\sqrt{c}|<\sqrt{a}<\sqrt{b}+\sqrt{c}$
so $\sqrt{a},\sqrt{b},\sqrt{c}$ are sides of a triangle.
WLOG assume $x\ge y\ge z$ .
We have $x(y-z)^2+z(x-y)^2-y(z-x)^2=(x-y)(z-y)(x+z)\le0$ , so $\sqrt{a}+\sqrt{c}\le\sqrt{b}$ , a contradiction.

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kuing

#3 h Sep 18, 2010, 4:20 PM • 3 Y

Y by Carl2015, Adventure10, Mango247

lssl wrote:

Let $x,y,z$ be positive reals such that $\sqrt{a}=x(y-z)^2$ , $\sqrt{b}=y(z-x)^2$ and $\sqrt{c}=z(x-y)^2$ . Prove that
$a^2+b^2+c^2 \geq 2(ab+bc+ca)$

note that:

$a^2 + b^2 + c^2 - 2\left( {ab + bc + ca} \right) = - \left( {\sqrt a + \sqrt b + \sqrt c } \right)\prod\limits_{cyc} {\left( {\sqrt a + \sqrt b - \sqrt c } \right)}$

but

$\sqrt a + \sqrt b - \sqrt c = x \left( {y - z} \right)^2 + y\left( {z - x} \right)^2 - z\left( {x - y} \right)^2$ $= - \left( {x + y} \right)\left( {y - z} \right)\left( {z - x} \right)$

so

$a^2 + b^2 + c^2 - 2\left( {ab + bc + ca} \right) = - \left( {\sqrt a + \sqrt b + \sqrt c } \right)\prod\limits_{cyc} {\left( { - \left( {x + y} \right)\left( {y - z} \right)\left( {z - x} \right)} \right)}$

$= \left( {\sqrt a + \sqrt b + \sqrt c } \right)\prod\limits_{cyc} {\left( {\left( {x + y} \right)\left( {x - y} \right)^2 } \right)} \ge 0$

done.

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jgnr

#4 h Sep 19, 2010, 12:05 PM • 1 Y

Y by Adventure10

In fact, the inequality is equivalent to $((a-b)(b-c)(c-a))^2(a+b)(b+c)(c+a)(a^2b+ab^2+b^2c+bc^2+c^2a+ca^2-6abc)\ge0$ .

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sqing

#5 h Nov 12, 2021, 1:17 PM

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Let $x,y,z\geq 0, a=x^2(y-z)^4,b=y^2(z-x)^4,c=z^2(x-y)^4$ . Prove that $\left( a+b+c \right)^3\geq\frac{9}{2}\left( a\sqrt{b}+b\sqrt{c}+c\sqrt{a} \right)^2$ 2021 Beijing University Golden Autumn Camp D2 P1
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This post has been edited 1 time. Last edited by sqing, Nov 12, 2021, 1:22 PM

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