If $xy+yz+xz=1$

[nirvanacs]

Let $x,y,z$ are positive numbers ,and such that
$xy+yz+xz=1$ prove that
$$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge 4(x+y+z)\left(\dfrac{x^2}{1+x^2}+\dfrac{y^2}{1+y^2 }+\dfrac{z^2}{1+z^2}\right)$$
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[blackholes.]

Nguyên văn bởi nirvanacs

Let $x,y,z$ are positive numbers ,and such that
$xy+yz+xz=1$ prove that
$$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge 4(x+y+z)\left(\dfrac{x^2}{1+x^2}+\dfrac{y^2}{1+y^2 }+\dfrac{z^2}{1+z^2}\right)$$

Bất đẳng thức đã cho tương đương với
$\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )\left ( xy+yz+zx \right )\geq 4\left ( x+y+z \right )\left ( \sum \frac{x^{2}}{\left ( x+z \right )\left ( y+z \right )} \right )$
$\Leftrightarrow 2+\frac{x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}}{xyz\left ( x+y+z \right )}\geq 4\sum \frac{x^{2}}{\left ( x+z \right )\left ( y+z \right )}$
Sử dụng hằng đẳng thức $\sum \frac{x^{2}}{\left ( x+z \right )\left ( x+y \right )}= 1-\frac{2xyz}{\left ( x+y\right )\left ( y+z \right )\left ( z+x \right )}$ ,bất đẳng thức được viết lại:
$\frac{x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}}{xyz\left ( x+y+z \right )}+\frac{8\left ( xy \right )\left ( yz \right )\left ( zx \right )}{\left ( xy+yz \right )\left ( yz+zx \right )\left ( zx+xy \right )}\geq 2$. Đây là một bất đẳng thức quen thuộc
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[nirvanacs]

Thank you, why
$\frac{x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}}{xyz\left ( x+y+z \right )}+\frac{8\left ( xy \right )\left ( yz \right )\left ( zx \right )}{\left ( xy+yz \right )\left ( yz+zx \right )\left ( zx+xy \right )}\geq 2$.
??

and I have this methods:
use this Schur inequality we have
$$\sum(a^2b+a^2c)(a-b)(a-c)\ge 0,a,b,c\ge 0$$
then we have
$$\Longrightarrow (\sum a)^2\prod (a+b)\ge 4\sum ab\sum ab(a+b)$$
or
$$(\sum a)^2\ge 4\sum ab\sum\dfrac{ab}{(a+c)(b+c)}$$
let $a=yz,b=zx, c=xy$
then we have
$$(\sum xy)^2\ge 4xyz\sum x\sum\dfrac{x^2}{(x+y)(x+z)}$$
use
$xy+yz+xz=1$ then we have
$$\sum\dfrac{1}{x}\ge 4xyz\sum x\sum\dfrac{x^2}{(x+y)(x+z)}$$
By done!


can you someone can use $C-S$ or $AM-GM$ inequality solve it? Thank you.
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[blackholes.]

Nguyên văn bởi nirvanacs

Thank you, why
$\frac{x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2}}{xyz\left ( x+y+z \right )}+\frac{8\left ( xy \right )\left ( yz \right )\left ( zx \right )}{\left ( xy+yz \right )\left ( yz+zx \right )\left ( zx+xy \right )}\geq 2$.
??

It is an useful inequality,you can prove it by SOS or SS method
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[nirvanacs]

Nguyên văn bởi blackholes.

It is an useful inequality,you can prove it by SOS or SS method

Oh, I forgot! It is well kown that
$$\dfrac{a^2+b^2+c^2}{ab+bc+ac}+\dfrac{8abc}{(a+b) (b+c)(c+a)}\ge 2 !$$
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