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sqing

#1 h Mar 31, 2025, 1:42 PM

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Let $a,b,c> 0$ and $abc=1$ . Prove that
$\frac1{a^2+a+k}+\frac1{b^2+b+k}+\frac1{c^2+c+k}\geq \frac{3}{k+2}$ Where $0<k \leq 1.$

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#2 h Apr 1, 2025, 6:30 AM

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sqing wrote:

Let $a,b,c> 0$ and $abc=1$ . Prove that
$\frac1{a^2+a+k}+\frac1{b^2+b+k}+\frac1{c^2+c+k}\geq \frac{3}{k+2}$ Where $0<k \leq 1.$

Let $a, b, c = \frac{x}{y}, \frac{y}{z}, \frac{z}{x}$ ; then the inequality is equivalent to

$\sum \frac{y^2}{x^2 + xy + ky^2} \ge \frac{3}{k+2}.$
By the Cauchy–Schwarz inequality,

$\sum \frac{y^2}{x^2+xy+ky^2} \ge \frac{\left(\sum y(y+z)\right)^2}{\sum (y+z)^2\left(x^2+xy+ky^2\right)}.$
It suffices to prove that

$(k+2)\left(\sum y(y+z)\right)^2 \ge 3\sum (y+z)^2\left(x^2+xy+ky^2\right).$
This is equivalent to

$3\sum xy(y-z)^2 + (1-k)\sum \left(x(x+y)-y(y+z)\right)^2 \ge 0,$
which is obviously true.

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sqing

#3 h Apr 1, 2025, 12:28 PM

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Very nice.Thank lbh_qys.

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#4 h Apr 1, 2025, 12:57 PM • 1 Y

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sqing wrote:

Let $a,b,c> 0$ and $abc=1$ . Prove that
$\frac1{a^2+a+k}+\frac1{b^2+b+k}+\frac1{c^2+c+k}\geq \frac{3}{k+2}$ Where $0<k \leq 1.$

Let $f(x)=\frac1{x^2+x+k}$ ==> $f'(x)=-\frac{2x+1}{(x^2+x+k)^2}$ ==> $f''(x)>0$ use Jensen : $\frac1{a^2+a+k}+\frac1{b^2+b+k}+\frac1{c^2+c+k}\geq 3f(\frac{\sum a}{3}) \geq 3f(1)=\frac{3}{k+2}$

This post has been edited 1 time. Last edited by SunnyEvan, Apr 1, 2025, 10:25 PM

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sqing

#5 h Apr 1, 2025, 1:24 PM

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Good.Thanks.

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sqing

#6 h Apr 1, 2025, 1:52 PM

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Let $a,b,c> 0$ and $a+b+c=3$ . Prove that
$\frac{a}{a^2+b+c+k}+\frac{b}{b^2+c+a+k}+\frac{c}{c^2+a+b+k} \leq \frac{3}{k+3}$ Where $k \geq 0.$

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#7 h Apr 1, 2025, 1:57 PM

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Jensen

:-D

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#8 h Apr 1, 2025, 10:53 PM

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sqing wrote:

Let $a,b,c> 0$ and $a+b+c=3$ . Prove that
$\frac{a}{a^2+b+c+k}+\frac{b}{b^2+c+a+k}+\frac{c}{c^2+a+b+k} \leq \frac{3}{k+3}$ Where $k \geq 0.$

$\frac{a}{a^2+b+c+k}+\frac{b}{b^2+c+a+k}+\frac{c}{c^2+a+b+k} = \frac{a}{a^2-a+3+k}+\frac{b}{b^2-b+3+k}+\frac{c}{c^2-c+3+k}$ Let $f(x)=\frac{x}{x^2-x+3+k}$
===> $f''(x)<0$ $\frac{a}{a^2-a+3+k}+\frac{b}{b^2-b+3+k}+\frac{c}{c^2-c+3+k} \leq3f(1)=\frac{3}{k+3}$

This post has been edited 2 times. Last edited by SunnyEvan, Apr 1, 2025, 10:53 PM

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sqing

#9 h Yesterday at 3:06 AM

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Thanks.
Let $a,b,c> 0$ and $a+b+c=3$ . Then
$\frac{a}{a^2+b+c}+\frac{b}{b^2+c+a}+\frac{c}{c^2+a+b} \leq \frac{a}{2a-1+b+c}+\frac{b}{2b-1+c+a}+\frac{c}{2c-1+a+b}$ $= \frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2} =3-2(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2})\leq 1$ $\frac{a}{a^2+b+c+2}+\frac{b}{b^2+c+a+2}+\frac{c}{c^2+a+b+2} \leq \frac{a}{2a+b+c+1}+\frac{b}{2b+c+a+1}+\frac{c}{2c+a+b+1}$ $= \frac{a}{a+4}+\frac{b}{b+4}+\frac{c}{c+4} \leq\frac{3}{5}$

This post has been edited 1 time. Last edited by sqing, Yesterday at 8:01 AM

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