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41908 posts

sqing

#1 h Apr 13, 2025, 10:16 AM • 1 Y

Y by cubres

Let $a,b\geq 0$ and $\frac{1}{a^2+b}+\frac{1}{b^2+a}=1.$ Prove that
$a^2+b^2-a-b\leq 1$ $a^3+b^3-a-b\leq \frac{3+\sqrt 5}{2}$ $a^3+b^3-a^2-b^2\leq \frac{1+\sqrt 5}{2}$

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#3 h Apr 13, 2025, 10:47 AM

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sqing wrote:

Let $a,b\geq 0$ and $\frac{1}{a^2+b}+\frac{1}{b^2+a}=1.$ Prove that
$a^2+b^2-a-b\leq 1$ $a^3+b^3-a-b\leq \frac{3+\sqrt 5}{2}$ $a^3+b^3-a^2-b^2\leq \frac{1+\sqrt 5}{2}$

$\frac{1}{a^2+b}+\frac{1}{b^2+a}=1 \iff a^3+b^3\in[2,2+\sqrt 5] , a^2+b^2 \in[1,\frac{3+\sqrt5}{2}] , a+b \in[\frac{1+\sqrt5}{2},2] , ab\in[0,1]$

This post has been edited 1 time. Last edited by SunnyEvan, Apr 13, 2025, 10:50 AM

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41908 posts

sqing

#4 h Apr 13, 2025, 10:52 AM

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Let $a,b\geq 0$ and $\frac{a+1}{a^2+b}+\frac{b+1}{b^2+a}=1.$ Prove that
$a+b+ab\leq 8$ $a+b-ab\leq\sqrt{2}+1$

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41908 posts

sqing

#5 h Apr 13, 2025, 10:53 AM

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Nice.Thank SunnyEvan.

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#6 h Apr 13, 2025, 10:59 AM

Y by

sqing wrote:

Let $a,b\geq 0$ and $\frac{a+1}{a^2+b}+\frac{b+1}{b^2+a}=1.$ Prove that
$a+b+ab\leq 8$ $a+b-ab\leq\sqrt{2}+1$

$\frac{a+1}{a^2+b}+\frac{b+1}{b^2+a}=1 \iff a+b \in[1+\sqrt 2,4] , ab \in[0,4]$

This post has been edited 1 time. Last edited by SunnyEvan, Apr 13, 2025, 11:00 AM

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sqing

#7 h Apr 13, 2025, 11:40 AM

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Let $a,b\geq 0$ and $\frac{a+1}{a^2+b}+\frac{b+2}{b^2+a}=1.$ Prove that
$a+2b+ab\leq 5+2\sqrt{13}$ $a+2b-ab\leq 2(1+\sqrt{3})$ $a+b-ab\leq \frac{3+\sqrt{13}}{2}$

This post has been edited 1 time. Last edited by sqing, Apr 13, 2025, 11:46 AM

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41908 posts

sqing

#8 h Apr 13, 2025, 11:41 AM

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Nice.Thank SunnyEvan.

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#9 h Apr 13, 2025, 2:59 PM

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sqing wrote:

Let $a,b\geq 0$ and $\frac{a+1}{a^2+b}+\frac{b+2}{b^2+a}=1.$ Prove that
$a+2b+ab\leq 5+2\sqrt{13}$ $a+2b-ab\leq 2(1+\sqrt{3})$ $a+b-ab\leq \frac{3+\sqrt{13}}{2}$

$\frac{a+1}{a^2+b}+\frac{b+2}{b^2+a}=1 \iff a+2b \in[3+\sqrt {13},\frac{3+3\sqrt{13}}{2}] , ab \in[0,\frac{7+\sqrt{13}}{2}]$

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sqing

#10 h Apr 13, 2025, 3:00 PM

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Very nice.Thank SunnyEvan.

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