inequality (my own)

[daogiauvang]

i wait to see a nice solution, Let $a,b,c$ be non-negative real numbers
$24(x^2-xy+y^2)^2(y^2-yz+z^2)^2(z^2-zx+x^2)^2 \geq xyz(x^2z+y^2x+z^2y)^3-3x^4y^4z^4$

[arqady]

daogiauvang писал(а):

i wait to see a nice solution, Let $a,b,c$ be non-negative real numbers
$24(x^2-xy+y^2)^2(y^2-yz+z^2)^2(z^2-zx+x^2)^2 \geq xyz(x^2z+y^2x+z^2y)^3-3x^4y^4z^4$

$$24\prod_{cyc}(x^2-xy+z^2)^2\geq24\prod_{cyc}{\frac{x^4+y^4}{2}=\sum_{cyc}(3x^8y^4+3x^8z^4+2x^4y^4z^4).$$

Поэтому остаётся доказать, что

$$3(x^4+y^4+z^4)(x^4y^4+x^4z^4+y^4z^4)\geq xyz(x^2z+y^2x+z^2y)^3.$$

Но по неравенству Гёлдера получаем:

$$3(x^4+y^4+z^4)(x^4y^4+x^4z^4+y^4z^4)=(1+1+1)(x^4+y^4+z^4)(x^4z^4+y^4x^4+z^4y^4)\geq\left(\sqrt[3]{x^8z^4}+\sqrt[3]{y^8x^4}+\sqrt[3]{z^8y^4}\right)^3.$$

То бишь осталось доказать, что

$$\sqrt[3]{x^8z^4}+\sqrt[3]{y^8x^4}+\sqrt[3]{z^8y^4}\geq\sqrt[3]{xyz}(x^2z+y^2x+z^2y)$$

то есть

$$a^8c^4+b^8a^4+c^8b^4\geq \sum_{cyc}a^7c^4b,$$

где

$$x=a^3,$$

$$y=b^3$$

и

$$z=c^3,$$

а это следует из следующего неравенства

$$10a^8c^4+b^8a^4+c^8b^4\geq12a^7c^4b,$$

которое верно по AM-GM.

[daogiauvang]

браво arqady.
My solution is same ur.
U're very clever when u solve inequality.

[arqady]

Thank you, daogiauvang!
A gift for you.
Let $a,$ $b$ and $c$ are non-negative numbers such that $a^3+b^3+c^3=3.$ Prove that

$$(a+b^2c^2)(b+a^2c^2)(c+a^2b^2)\geq8a^2b^2c^2$$

[daogiauvang]

arqady писал(а):

Thank you, daogiauvang!
A gift for you.
Let $a,$ $b$ and $c$ are non-negative numbers such that $a^3+b^3+c^3=3.$ Prove that

$$(a+b^2c^2)(b+a^2c^2)(c+a^2b^2)\geq8a^2b^2c^2$$

i have a solution:
Let $abc=r \leq 1$
The inequality is equavilent to:
$a^3 b^3 + b^3 c^3 + c^3 a^3 + abc + a^4 b^4 c^4 \ge 5a^2 b^2 c^2$
Note that:$(a^3 b^3 + b^3 c^3 + c^3 a^3)^2\geq 3(a^3+b^3+c^3)a^3b^3c^3$hence $a^3 b^3 + b^3 c^3 + c^3 a^3 \geq 3\sqrt{a^3b^3c^3}=3\sqrt{r^3}$
Hence:
$3\sqrt{r^3}+r+r^4 \geq 5 \sqrt[5]{r^{19/2}} \geq 5 \sqrt[5]{r^{10}} \leftrightarrow r \leq 1$
The inequality is solved.

[arqady]

Всё правильно, daogiauvang!
А если $a^4+b^4+c^4=3$ $?$