General inequality 1

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General inequality 1

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#1 h Apr 19, 2016, 9:21 AM • 3 Y

Y by ali3985, Adventure10, Mango247

If $x, y$ and $z$ are positive real number and $k > 1$ such that $(x+y+z)\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )=9+\frac{8}{k^2-1},$ then $\frac{2(k+1)^3+(k-1)^3}{(k-1)(k+1)^2} \leqslant \frac{x^3+y^3+z^3}{xyz} \leqslant \frac{2(k-1)^3+(k+1)^3}{(k+1)(k-1)^2}.$ Click to reveal hidden text

This post has been edited 6 times. Last edited by Nguyenhuyen_AG, May 23, 2016, 1:07 PM

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#2 h Apr 19, 2016, 2:43 PM • 2 Y

Y by Adventure10, Mango247

phép đặt
$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=x$
$\frac{b}{a}+\frac{c}{b}+\frac{a}{c}=y$

giúp chúng ta có kết quả ,có đúng không ạ !

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Quantum_fluctuations

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Quantum_fluctuations

#3 h Nov 18, 2020, 2:28 AM

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Nguyenhuyen_AG wrote:

If $x, y$ and $z$ are positive real number and $k > 1$ such that $(x+y+z)\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )=9+\frac{8}{k^2-1},$ then $\frac{2(k+1)^3+(k-1)^3}{(k-1)(k+1)^2} \leqslant \frac{x^3+y^3+z^3}{xyz} \leqslant \frac{2(k-1)^3+(k+1)^3}{(k+1)(k-1)^2}.$ Click to reveal hidden text

How to prove it?

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#4 h Nov 18, 2020, 9:49 PM • 3 Y

Y by Mango247, Mango247, Mango247

Nguyenhuyen_AG wrote:

If $x, y$ and $z$ are positive real number and $k > 1$ such that $(x+y+z)\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )=9+\frac{8}{k^2-1},$ then $\frac{2(k+1)^3+(k-1)^3}{(k-1)(k+1)^2} \leqslant \frac{x^3+y^3+z^3}{xyz} \leqslant \frac{2(k-1)^3+(k+1)^3}{(k+1)(k-1)^2}.$ Click to reveal hidden text

similar

Let $x, y, z,$ be positive real numbers such that $k>1$ $(x+y+z)\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )=9+\frac{8}{k^2-1},$ Prove that $\frac{2(k+1)^3+(k-1)^3}{(k-1)(k+1)^2} \leqslant \frac{x^3y^3+y^3z^3+z^3x^3}{x^2y^2z^2} \leqslant \frac{2(k-1)^3+(k+1)^3}{(k+1)(k-1)^2}.$

This post has been edited 3 times. Last edited by Auditor, Nov 18, 2020, 9:51 PM

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#5 h Nov 13, 2021, 1:46 PM

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Nguyenhuyen_AG wrote:

If $x, y$ and $z$ are positive real number and $k > 1$ such that $(x+y+z)\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right )=9+\frac{8}{k^2-1},$ then $\frac{2(k+1)^3+(k-1)^3}{(k-1)(k+1)^2} \leqslant \frac{x^3+y^3+z^3}{xyz} \leqslant \frac{2(k-1)^3+(k+1)^3}{(k+1)(k-1)^2}.$ Click to reveal hidden text

how to solve it ?

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3377 posts

mudok

#6 h Nov 14, 2021, 1:31 PM

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Let $x+y+z=3, xy+yz+zx=3(1-t^2), xyz=r$ , where $t\in [0,1)$ .

The condition gives us: $9(1-t^2) = \frac{r(9k^2-1)}{k^2-1}$ .

From the well-known result $(1+t)^2(1-2t)\le r\le (1+2t)(1-t)^2$ , we get:

$9(1-t^2) \ge \frac{(1+t)^2(1-2t)(9k^2-1)}{k^2-1} \Rightarrow \ \ t\ge \frac{2}{3k+1}$

$9(1-t^2) \le \frac{(1-t)^2(1+2t)(9k^2-1)}{k^2-1} \Rightarrow \ \ t\le \frac{2}{3k-1}$

We need to prove:

$\frac{2(k+1)^3+(k-1)^3}{(k-1)(k+1)^2} - 3 \le \frac{x^3+y^3+z^3}{xyz} -3\le \frac{2(k-1)^3+(k+1)^3}{(k+1)(k-1)^2} -3$

$\iff \frac{4(3k+1)}{(k-1)(k+1)^2}\le \frac{3t^2(9k^2-1)}{(1-t^2)(k^2-1)}\le \frac{4(3k-1)}{(k+1)(k-1)^2}$

$\iff \frac{4}{3(k+1)(3k-1)}\le \frac{t^2}{1-t^2}\le \frac{4}{3(k-1)(3k+1)}$

which is true due to $\frac{2}{3k+1}\le t\le \frac{2}{3k-1}$

This post has been edited 4 times. Last edited by mudok, Nov 14, 2021, 1:56 PM

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