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sqing

#1 h Apr 9, 2025, 4:41 AM

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Let $a,b$ be real numbers . Prove that
$-\frac{1}{12}\leq \frac{ab+a+b+1}{(a^2+3)(b^2+3)}\leq \frac{1}{4}$ $-\frac{5}{96}\leq \frac{ab+a+b+2}{(a^2+4)(b^2+4)}\leq \frac{1}{5}$ $-\frac{1}{16}\leq \frac{ab+a+b+1}{(a^2+4)(b^2+4)}\leq \frac{3+\sqrt 5}{32}$ $-\frac{1}{18}\leq \frac{ab+a+b+3}{(a^2+3)(b^2+3)}\leq \frac{2+\sqrt[3] 2+\sqrt[3] {4}}{12}$

This post has been edited 1 time. Last edited by sqing, Apr 9, 2025, 4:41 AM

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#2 h Apr 9, 2025, 5:16 AM • 1 Y

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sqing wrote:

Let $a,b$ be real numbers . Prove that
$-\frac{1}{12}\leq \frac{ab+a+b+1}{(a^2+3)(b^2+3)}\leq \frac{1}{4}$

Because $-\frac{1}{6} \leq \frac{x+1}{x^2+3} \leq \frac{1}{2}$

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#3 h Apr 9, 2025, 5:42 AM

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sqing wrote:

Let $a,b$ be real numbers . Prove that
$\frac{ab+a+b+2}{(a^2+4)(b^2+4)}\leq \frac{1}{5}$

According to the AM-GM inequality

$\frac{(a+1)^2}{(a^2+4)^2} + \frac{(b+1)^2}{(b^2+4)^2} \ge \frac{2(a+1)(b+1)}{(a^2+4)(b^2+4)}$
$\frac{1}{(a^2+4)^2} + \frac{1}{(b^2+4)^2} \ge \frac{2}{(a^2+4)(b^2+4)}$
whence

$\frac{2(ab+a+b+2)}{(a^2+4)(b^2+4)} \le \frac{(a+1)^2+1}{(a^2+4)^2} + \frac{(b+1)^2+1}{(b^2+4)^2}$
It suffices to prove that

$\frac{(x+1)^2+1}{(x^2+4)^2} \le \frac{1}{5}$

This post has been edited 1 time. Last edited by lbh_qys, Apr 9, 2025, 6:44 AM

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#4 h Apr 9, 2025, 6:54 AM

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sqing wrote:

Let $a,b$ be real numbers . Prove that
$-\frac{5}{96}\leq \frac{ab+a+b+2}{(a^2+4)(b^2+4)}$ $-\frac{1}{18}\leq \frac{ab+a+b+3}{(a^2+3)(b^2+3)}$

Let $a,b$ be real numbers. Prove that
$\frac{ab + a + b + \lambda}{ (a^2+\mu)(b^2+\mu)} \ge - \frac{\mu + 1}{4\mu(\lambda + \mu)}$
where $\lambda, \mu > 0$ and $\lambda,\mu \in \mathbb{R}$ .

This post has been edited 3 times. Last edited by lbh_qys, Apr 9, 2025, 7:08 AM

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#5 h Apr 9, 2025, 7:25 AM

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lbh_qys wrote:

sqing wrote:

Let $a,b$ be real numbers . Prove that
$-\frac{5}{96}\leq \frac{ab+a+b+2}{(a^2+4)(b^2+4)}$ $-\frac{1}{18}\leq \frac{ab+a+b+3}{(a^2+3)(b^2+3)}$

Let $a,b$ be real numbers. Prove that
$\frac{ab + a + b + \lambda}{ (a^2+\mu)(b^2+\mu)} \ge - \frac{\mu + 1}{4\mu(\lambda + \mu)}$
where $\lambda, \mu > 0$ and $\lambda,\mu \in \mathbb{R}$ .

Since
$4xy \ge - (x-y)^2,$ we have
$4(ab+a+b+\lambda)(\mu+\lambda)\ge -\Bigl((ab+a+b+\lambda)-(\mu+\lambda)\Bigr)^2 = -(ab+a+b-\mu)^2.$
According to the Cauchy--Schwarz inequality,
$(\mu+1)(a^2+\mu)(b^2+\mu)=(\mu+1)\Bigl((ab-\mu)^2+\mu(a+b)^2\Bigr) \ge \Bigl(\sqrt{\mu}(ab-\mu)+\sqrt{\mu}(a+b)\Bigr)^2 = \mu(ab+a+b-\mu)^2.$
Thus
$4(ab+a+b+\lambda)(\mu+\lambda)\mu + (\mu+1)(a^2+\mu)(b^2+\mu) \ge -(ab+a+b-\mu)^2 \mu + \mu(ab+a+b-\mu)^2 = 0$ which is
$\frac{ab+a+b+\lambda}{(a^2+\mu)(b^2+\mu)}\ge -\frac{\mu+1}{4\mu(\lambda+\mu)}.$

This post has been edited 1 time. Last edited by lbh_qys, Apr 9, 2025, 7:27 AM

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#6 h Apr 9, 2025, 8:08 AM

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Let $a,b$ be real numbers. Prove that
$\frac{ab + a + b - \lambda}{ (a^2+\mu)(b^2+\mu)} \le \frac{\mu + 1}{4\mu(\lambda - \mu)}$
where $\mu > 0, \lambda \geq 2(\mu+1)^2$ and $\lambda,\mu \in \mathbb{R}$ .

$\frac{ab + a + b - 32}{ (a^2+3)(b^2+3)} \le \frac{1}{87}$
$\frac{ab + a + b - 50}{ (a^2+4)(b^2+4)} \le \frac{5}{736}$

This post has been edited 1 time. Last edited by lbh_qys, Apr 9, 2025, 8:09 AM

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sqing

#7 h Apr 9, 2025, 9:19 AM

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Very very nice.Thank lbh_qys.

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sqing

#8 h Apr 9, 2025, 9:31 AM

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Let $a,b$ be real numbers. Prove that
$\frac{ab + a + b - 29}{ (a^2+3)(b^2+3)} \le \frac{1}{78}$ $\frac{ab + a + b - 31}{ (a^2+3)(b^2+3)} \le \frac{1}{84}$ $\frac{ab + a + b - 30}{ (a^2+3)(b^2+3)} \le \frac{1}{81}$ $\frac{ab + a + b - 49}{ (a^2+4)(b^2+4)} \le \frac{1}{144}$ $\frac{ab + a + b - 48}{ (a^2+4)(b^2+4)} \le \frac{5}{704}$ $\frac{ab + a + b - 47}{ (a^2+4)(b^2+4)} \le \frac{5}{688}$

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#9 h Apr 9, 2025, 12:08 PM

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lbh_qys wrote:

Let $a,b$ be real numbers. Prove that
$\frac{ab + a + b - \lambda}{ (a^2+\mu)(b^2+\mu)} \le \frac{\mu + 1}{4\mu(\lambda - \mu)}$
where $\mu > 0, \lambda \geq 2(\mu+1)^2$ and $\lambda,\mu \in \mathbb{R}$ .

$\frac{ab + a + b - 32}{ (a^2+3)(b^2+3)} \le \frac{1}{87}$
$\frac{ab + a + b - 50}{ (a^2+4)(b^2+4)} \le \frac{5}{736}$

Since
$(x+y)^2 \geq 4xy$ $\frac{ab + a + b - \lambda}{ (a^2+\mu)(b^2+\mu)} \le \frac{\mu + 1}{4\mu(\lambda - \mu)} \iff 4\mu(\lambda - \mu)(ab + a + b - \lambda) \leq (a^2+\mu)(b^2+\mu)(\mu + 1)$ $4\mu(\lambda - \mu)(ab + a + b - \lambda) \leq \mu(ab+a+b-\mu)^2$ and $(\mu+1)(a^2+\mu)(b^2+\mu) \geq \mu(ab+a+b-\mu)^2$ done.

This post has been edited 3 times. Last edited by SunnyEvan, Apr 9, 2025, 12:10 PM

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sqing

#10 h Apr 9, 2025, 12:26 PM

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Nice.Thanks.

This post has been edited 1 time. Last edited by sqing, Apr 9, 2025, 12:26 PM

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