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sqing

#1 h Apr 13, 2025, 2:42 AM

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Let $a,b\geq 0$ and $a^2+ab+b^2=2$ . Prove that
$(a+b-ab)\left( \frac{1}{a+1} + \frac{1}{b+1}\right)\leq 2$ $(a+b-ab)\left( \frac{a}{b+1} + \frac{2b}{a+2}\right)\leq 2$ $(a+b-ab)\left( \frac{a}{b+1} + \frac{2b}{a+1}\right)\leq 4$ Let $a,b$ be reals such that $a^2+b^2=2$ . Prove that
$(a+b)\left( \frac{1}{a+1} + \frac{1}{b+1}\right)= 2$ $(a+b)\left( \frac{a}{b+1} + \frac{b}{a+1}\right)=2$

This post has been edited 1 time. Last edited by sqing, Apr 13, 2025, 6:41 AM

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sqing

#2 h Apr 13, 2025, 2:44 AM

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Let $a,b\geq 0$ and $a^2+b^2=2$ . Prove that
$(a+b)\left( \frac{1}{a+1} + \frac{2}{b+1}\right)\leq 2+\sqrt 2$ $(a+b)\left( \frac{1}{a+1} + \frac{1}{b+2}\right)\leq 2\sqrt 2-1$ $(a+b)\left( \frac{a}{b+1} + \frac{2b}{a+1}\right)\leq 4$ $(a+b)\left( \frac{a}{b+1} + \frac{b}{a+2}\right)\leq 2$

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sqing

#3 h Apr 13, 2025, 3:08 AM

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Let $a,b,c\geq 0$ and $a^2+b^2+c^2=2$ . Prove that
$(a+b+c)\left( \frac{1}{a+2} + \frac{ 1}{ b+1}+ \frac{ 1}{ c+2}\right)\leq \frac{10}{3}$

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#4 h Apr 13, 2025, 5:48 AM • 1 Y

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sqing wrote:

Let $a,b$ be reals such that $a^2+b^2=2$ . Prove that
$(a+b)\left( \frac{1}{a+1} + \frac{1}{b+1}\right)\leq 2$ $(a+b)\left( \frac{a}{b+1} + \frac{b}{a+1}\right)\leq 2$

$(a+b)\left( \frac{1}{a+1} + \frac{1}{b+1}\right)= 2$ $(a+b)\left( \frac{a}{b+1} + \frac{b}{a+1}\right)= 2$

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sqing

#5 h Apr 13, 2025, 6:25 AM

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Good.Thank SunnyEvan.

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#6 h Apr 13, 2025, 7:06 AM

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Let $a,b\geq 0$ and $a^2+b^2=2$ . Prove that
$(a+b)\left( \frac{a}{b+1} + \frac{2b}{a+1}\right)\leq 4$

sol

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sqing

#7 h Apr 13, 2025, 7:44 AM

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Thanks.

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#9 h Apr 13, 2025, 10:14 AM

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sqing wrote:

Let $a,b\geq 0$ and $a^2+ab+b^2=2$ . Prove that
$(a+b-ab)\left( \frac{1}{a+1} + \frac{1}{b+1}\right)\leq 2$ $(a+b-ab)\left( \frac{a}{b+1} + \frac{2b}{a+2}\right)\leq 2$ $(a+b-ab)\left( \frac{a}{b+1} + \frac{2b}{a+1}\right)\leq 4$ Let $a,b$ be reals such that $a^2+b^2=2$ . Prove that
$(a+b)\left( \frac{1}{a+1} + \frac{1}{b+1}\right)= 2$ $(a+b)\left( \frac{a}{b+1} + \frac{b}{a+1}\right)=2$

Let $u=a+b$ and $v=ab$
so : $a^2+ab+b^2=2 \iff u^2=v+2 \rightarrow v\in[0,\frac{8}{3}]$ and $a+b=\sqrt{ab+2} \rightarrow a+b-ab \geq 0$ when $ab\leq 2$
$1.$ $(a+b-ab)\left( \frac{1}{a+1} + \frac{1}{b+1}\right)\leq 2 \iff (u-v)(\frac {u+2}{u+v+1}) \leq 2 \iff u^3+3u^2-2u-6 \geq 0 \iff u\geq \sqrt 2$ $2.$ when $a+b-ab\geq 0$
$(a+b-ab)\left( \frac{a}{b+1} + \frac{2b}{a+2}\right)\leq (a+b-ab)(a+b)\leq 2$ when $a+b-ab \leq 0$ , $(a+b-ab)( \frac{a}{b+1} + \frac{2b}{a+2}) \leq 0$
The third one is same.

This post has been edited 1 time. Last edited by SunnyEvan, Apr 13, 2025, 10:15 AM

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sqing

#10 h Apr 13, 2025, 10:22 AM

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sqing wrote:

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

Solution of DAVROS:
Let $u=a+b, v=ab$ then $u^2=3v+1, u^2\ge4v$ so $v=\frac{u^2-1}3, 1\le u\le2$ and $v\le1$

$\textbf{LHS} = (u-v)\left( \frac{a^2+b^2+a+b}{ab+a+b+1}-v\right) =(u-v)(1-v) \le \frac{(u+1-2v)^2}4$

$\textbf{LHS} \le \frac{(-2u^2 + 3u + 5)^2}{36} = \frac{\left(\frac{49}8-2\left(u-\frac34\right)^2\right)^2}{36}\le 1$ at $u=1, v=0 \implies (a,b) =\{0,1\}$

Very nice.Thank SunnyEvan.

This post has been edited 1 time. Last edited by sqing, Apr 14, 2025, 8:48 AM

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