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#1 h Apr 20, 2025, 9:39 AM

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Let $a,b,c \geq 0$ Prove that :
$\frac{3k}{k^4+k+1} \leq \frac{ka}{a+k^4b+kc}+\frac{kb}{b+k^4c+ka}+\frac{kc}{c+k^4a+kb} \leq k$ Where $k \geq 1$

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#4 h Apr 20, 2025, 11:43 AM

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Bump for it.

This post has been edited 1 time. Last edited by SunnyEvan, Apr 20, 2025, 12:27 PM

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#5 h Apr 20, 2025, 12:17 PM

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SunnyEvan wrote:

Let $a,b,c \geq 0$ Prove that :
$\frac{3k}{k^4+k+1} \leq \frac{ka}{a+k^4b+kc}+\frac{kb}{b+k^4c+ka}+\frac{kc}{c+k^4a+kb} \leq k$ Where $k \geq 1$

The Right side is easy:
$\frac{ka}{a+k^4b+kc}+\frac{kb}{b+k^4c+ka}+\frac{kc}{c+k^4a+kb} \leq \frac{ka}{a+b+c}+\frac{kb}{b+c+a}+\frac{kc}{c+a+b} =k$
For the Left side, you can use Cauchy-Schwarz, and let $t = \frac{a^2+b^2+c^2}{ab+bc+ca} \geq 1$ . and we only need to prove:
$\frac{(a+b+c)^2}{a^2+b^2+c^2+(k^4+k)(ab+bc+ca)} = \frac{t+2}{t+k^4+k}\geq \frac{3}{k^4+k+1}$ Or:
$\frac{(k-1)(t-1)(k^3+k^2+k+1)}{(k^4+k+1)(k^4+k+t)} \geq 0$ which is true.

This post has been edited 4 times. Last edited by GeoMorocco, Apr 20, 2025, 12:37 PM

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#6 h Apr 20, 2025, 12:20 PM

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Let $a,b,c \geq 0$ Prove that :
$\frac{a}{ka+k^2b+c}+\frac{b}{kb+k^2c+a}+\frac{c}{kc+k^2a+b} \leq \frac{1}{k} \leq \frac{a}{ka+(2k-1)b+c}+\frac{b}{kb+(2k-1)+a}+\frac{c}{kc+(2k-1)a+b}$ Where $k \geq 1$

This post has been edited 3 times. Last edited by SunnyEvan, Apr 25, 2025, 1:21 PM

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#7 h Apr 20, 2025, 12:35 PM

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GeoMorocco wrote:

SunnyEvan wrote:

Let $a,b,c \geq 0$ Prove that :
$\frac{3k}{k^4+k+1} \leq \frac{ka}{a+k^4b+kc}+\frac{kb}{b+k^4c+ka}+\frac{kc}{c+k^4a+kb} \leq k$ Where $k \geq 1$

The Right side is easy:
$\frac{ka}{a+k^4b+kc}+\frac{kb}{b+k^4c+ka}+\frac{kc}{c+k^4a+kb} \leq \frac{ka}{a+b+c}+\frac{kb}{b+c+a}+\frac{kc}{c+a+b} =k$
For the Left side, you can use Cauchy-Schwarz, and let $\frac{a^2+b^2+c^2}{ab+bc+ca} \geq 1$ . and we only need to prove:
$\frac{(a+b+c)^2}{a^2+b^2+c^2+(k^4+k)(ab+bc+ca)} = \frac{t+2}{t+k^4+k}\geq \frac{3}{k^4+k+1}$ which is true:
$\frac{(k-1)(t-1)(k^3+k^2+k+1)}{(k^4+k+1)(k^4+k+t)} \geq 0$ which is true.

Thanks for your help GeoMorocco.

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#8 h Apr 25, 2025, 1:22 PM

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SunnyEvan wrote:

Let $a,b,c \geq 0$ Prove that :
$\frac{a}{ka+k^2b+c}+\frac{b}{kb+k^2c+a}+\frac{c}{kc+k^2a+b} \leq \frac{1}{k} \leq \frac{a}{ka+(2k-1)b+c}+\frac{b}{kb+(2k-1)+a}+\frac{c}{kc+(2k-1)a+b}$ Where $k \geq 1$

How about this?

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