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sqing

#1 h Apr 2, 2025, 3:12 AM

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Let $a,b,c\geq 0$ and $a+b+c=3$ . Prove that
$\frac{a^2}{a^2+b+c+ \frac{3}{2}}+\frac{b^2}{b^2+c+a+\frac{3}{2}}+\frac{c^2}{c^2+a+b+\frac{3}{2}} \leq \frac{6}{7}$ Equality holds when $(a,b,c)=(0,\frac{3}{2},\frac{3}{2})$ or $(a,b,c)=(0,0,3) .$

This post has been edited 2 times. Last edited by sqing, Apr 2, 2025, 4:31 AM

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#2 h Apr 2, 2025, 3:15 AM

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Let $a,b,c\geq 0$ and $a+b+c=3$ . Prove that
$\frac{a^2}{a^2+b+c+k}+\frac{b^2}{b^2+c+a+k}+\frac{c^2}{c^2+a+b+k} \leq \frac{18}{4k+15}$ Where $0< k \leq \frac{3}{2}.$

This post has been edited 1 time. Last edited by sqing, Apr 2, 2025, 4:31 AM

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#3 h Apr 2, 2025, 4:22 AM

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sqing wrote:

Let $a,b,c> 0$ and $a+b+c=3$ . Prove that
$\frac{a^2}{a^2+b+c+ \frac{3}{2}}+\frac{b^2}{b^2+c+a+\frac{3}{2}}+\frac{c^2}{c^2+a+b+\frac{3}{2}} \leq \frac{6}{7}$ Equality holds when $(a,b,c)=(0,\frac{3}{2},\frac{3}{2})$ or $(a,b,c)=(0,0,3) .$

Based on the C-S inequality, we have
$\frac{3a^2}{a^2+b^2+c^2}+\frac{4a}{a+b+c}\ge\frac{(3a+4a)^2}{3(a^2+b^2+c^2)+4a(a+b+c)}=\frac{49a^2}{7a^2+3b^2+3c^2+4a(b+c)}\ge\frac{49a^2}{7a^2+3b^2+3c^2+4a(b+c)+6bc}=\frac{49}{6}\frac{a^2}{a^2+b+c+\frac{3}{2}}.$
Summing over, we obtain
$\frac{49}{6}\sum\frac{a^2}{a^2+b+c+\frac{3}{2}}\le3\sum\frac{a^2}{a^2+b^2+c^2}+4\sum\frac{a}{a+b+c}=7,$ and hence,
$\sum\frac{a^2}{a^2+b+c+\frac{3}{2}}\le\frac{6}{7}.$

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sqing

#4 h Apr 2, 2025, 4:33 AM

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Very nice.Thank lbh_qys.

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#5 h Apr 2, 2025, 4:37 AM

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sqing wrote:

Let $a,b,c\geq 0$ and $a+b+c=3$ . Prove that
$\frac{a^2}{a^2+b+c+k}+\frac{b^2}{b^2+c+a+k}+\frac{c^2}{c^2+a+b+k} \leq \frac{18}{4k+15}$ Where $0< k \leq \frac{3}{2}.$

According to the Cauchy–Schwarz inequality, we have

$\frac{(k+3)a^2}{a^2+b^2+c^2}+\frac{(2k+3)a}{a+b+c}+\left(\frac{3}{2}-k\right)\frac{a^2}{2a^2+bc} \ge \frac{\big[(k+3)a+(2k+3)a+\left(\frac{3}{2}-k\right)a\big]^2}{(k+3)(a^2+b^2+c^2)+(2k+3)a(a+b+c)+\left(\frac{3}{2}-k\right)(2a^2+bc)}$
$=\frac{(4k+15)^2}{4}\frac{a^2}{9a^2+3\Bigl(ab+ac+b^2+c^2+\frac{1}{2}bc\Bigr)+k\Bigl(a^2+b^2+c^2+2ab+2ac-bc\Bigr)}$
$\ge \frac{(4k+15)^2}{4}\frac{a^2}{9a^2+3\Bigl(ab+ac+b^2+c^2+2bc\Bigr)+k\Bigl(a^2+b^2+c^2+2ab+2ac+2bc\Bigr)}$
$=\frac{(4k+15)^2}{4}\frac{a^2}{9a^2+3(b+c)(a+b+c)+k(a+b+c)^2}=\frac{(4k+15)^2}{36}\frac{a^2}{a^2+b+c+k}.$
Summing over the cyclic permutations and applying $\sum \frac{a^2}{2a^2+bc}\leq 1$ , we obtain

$\frac{(4k+15)^2}{36}\sum \frac{a^2}{a^2+b+c+k}\leq (k+3)\sum \frac{a^2}{a^2+b^2+c^2}+(2k+3)\sum \frac{a}{a+b+c}+\left(\frac{3}{2}-k\right)\sum \frac{a^2}{2a^2+bc}$ $\leq k+3+2k+3+\left(\frac{3}{2}-k\right)=\frac{4k+15}{2}.$
Therefore,

$\sum \frac{a^2}{a^2+b+c+k}\leq \frac{18}{4k+15}.$

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sqing

#6 h Apr 2, 2025, 4:57 AM

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Very nice.Thank lbh_qys.

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sqing

#7 h Apr 2, 2025, 9:09 AM

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Let $a,b,c\geq 0$ and $a+b+c=3$ . Prove that
$\frac{ab}{c^2+a+b+k} +\frac{bc}{a^2+b+c+k}+\frac{ca}{b^2+c+a+k}\leq \frac{3}{k+3}$ $\frac{a+k}{a^2+b+c}+\frac{b+k}{b^2+c+a}+\frac{c+k}{c^2+a+b} \leq k+1$ Where $k\geq -\frac{3}{2}.$

This post has been edited 1 time. Last edited by sqing, Apr 2, 2025, 9:21 AM

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