Inequalities

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sqing

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#1 h Nov 14, 2024, 11:56 AM

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Let $a,b \geq 0 , a^2+b^2= 3$ and $a^4+b^4 = 7.$ Prove that $\frac{15\sqrt 5-11}{2}\leq a^5+2b^5\leq \frac{15\sqrt 5+11}{2}$ $15\sqrt 5-22 \leq a^5 +5b^5\leq 15\sqrt 5+22$

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#2 h Nov 14, 2024, 12:26 PM

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Let $a ,b>0$ and $(a+1)(2b+3)=4$ . Prove that $2ab+\frac{1}{ab}\geq 21+4\sqrt{3}$ Let $a ,b>0$ and $(a+1)(2b+3)=6$ . Prove that $2ab+\frac{1}{ab}\geq 11-\frac{14\sqrt{2}}{3}$

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DAVROS

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#3 h Nov 15, 2024, 7:03 AM

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sqing wrote:

Let $a ,b>0$ and $(a+1)(2b+3)=6$ . Prove that $2ab+\frac{1}{ab}\geq 11-\frac{14\sqrt{2}}{3}$

solution

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#4 h Nov 15, 2024, 7:25 AM

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Very very nice.Thank DAVROS.

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#5 h Nov 15, 2024, 7:36 AM

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Let $a,b,c\in [0,1]$ and $a+b+c=2 .$ Prove that
$\dfrac{1}{2}\leq \dfrac{1-a^2}{a+2}+\dfrac{1-b^2}{b+2}+\dfrac{1-c^2}{c+2} \leq \dfrac{5}{8}$ $\dfrac{14}{5}\leq\dfrac{2-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{2-c^2}{c+1}\leq3$ Let $a,b,c\in [0,1]$ and $a+b+c=1 .$ Prove that
$1\leq\dfrac{1-a^2}{a+2}+\dfrac{1-b^2}{b+2}+\dfrac{1-c^2}{c+2} \leq \dfrac{8}{7}$ $\dfrac{17}{4}\leq\dfrac{2-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{2-c^2}{c+1} \leq \dfrac{9}{2}$

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sqing

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#6 h Nov 15, 2024, 7:55 AM

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Let $a,b,c\in [0,2]$ and $a+b+c=3 .$ Prove that
$1+\dfrac{2\sqrt 2}{3}\leq \dfrac{2-a^2}{a+1}+\dfrac{3-b^2}{b+1}+\dfrac{2-c^2}{c+1}\leq \dfrac{17}{6}$ $\dfrac{9+4\sqrt 2}{6}\leq \dfrac{3-a^2}{a+1}+\dfrac{2-b^2}{b+1}+\dfrac{3-c^2}{c+1} \leq \dfrac{10}{3}$

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sqing

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#7 h Nov 23, 2024, 1:44 AM

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Let $a,b,c \in [\frac{1}{2},2] .$ Prove that
$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\leq \frac{27}{2}$ $(a+2b+c)\left(\frac{1}{a}+\frac{2}{b}+\frac{1}{c}\right)\leq 25$ $(a+2b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{2}{c}\right)\leq \frac{65}{2}$ $(a+ b^2+ c)\left(\frac{1}{a}+\frac{ 1}{b}+\frac{1}{c}\right) \leq \frac{45}{2}$ $(a+ b^2+ c)\left(\frac{1}{a}+\frac{ 1}{b}+\frac{1}{c^2}\right) \leq \frac{65}{2}$ $(a+ b^2+ c)\left(\frac{1}{a}+\frac{ 1}{b^2}+\frac{1}{c}\right) \leq \frac{85}{4}$

This post has been edited 1 time. Last edited by sqing, Nov 23, 2024, 1:48 AM

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DAVROS

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#8 h Nov 23, 2024, 9:58 AM

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sqing wrote:

Let $a,b,c\in [0,1]$ and $a+b+c=2 .$ Prove that $\dfrac{1}{2}\leq \dfrac{1-a^2}{a+2}+\dfrac{1-b^2}{b+2}+\dfrac{1-c^2}{c+2} \leq \dfrac{5}{8}$

solution

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#9 h Nov 23, 2024, 10:28 AM

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sqing wrote:

Let $a,b,c \in [\frac{1}{2},2] .$ Prove that $(a+ b^2+ c)\left(\frac{1}{a}+\frac{ 1}{b^2}+\frac{1}{c}\right) \leq \frac{85}{4}$

solution

This post has been edited 1 time. Last edited by DAVROS, Nov 23, 2024, 10:29 AM

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#10 h Nov 23, 2024, 12:04 PM

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Very very nice.Thank DAVROS.

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sqing

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#11 h Nov 23, 2024, 1:30 PM

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Let $a,b,c \in[1, 2]$ and $a^2+b^2+c^2=11$ . Prove that
$2b+3c \le \frac{8}{\sqrt{3}} a$ Let $a,b,c \in[1, 3]$ and $a^2+b^2+c^2=14$ . Prove that
$2b+5c \le 19 \\\ a$ $5b+6c \le \sqrt{793}\\\ a$

This post has been edited 1 time. Last edited by sqing, Nov 23, 2024, 1:41 PM

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#12 h Nov 23, 2024, 2:38 PM

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Let $a,b,c>0$ and $a^3+b^3+c^3+1=4abc$ . Prove that
$(2-a)(2-b)(2-c) \leq abc$ $(3-a)(3-b)(3-c) \leq 8abc$ $(4-3a)(4- 3b )(4-3c ) \leq abc$

This post has been edited 1 time. Last edited by sqing, Nov 24, 2024, 11:08 AM

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#13 h Nov 25, 2024, 1:32 AM

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Let $a,b,c$ be reals such that $a^2+2b^2= 9 ,3b^2+4c^2 = 48$ and $5c^2+6a^2 = 51.$ Prove that $-22\leq a+3b+5c\leq 22$ $-10\leq 3a-2b+c\leq 10$

This post has been edited 2 times. Last edited by sqing, Nov 26, 2024, 1:26 AM

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lbh_qys

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#14 h Nov 25, 2024, 6:41 AM

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sqing wrote:

Let $a,b,c>0$ and $a^3+b^3+c^3+1=4abc$ . Prove that
$(2-a)(2-b)(2-c) \leq abc$

By AM-GM, $4abc = a^3 + b^3 + c^3 + 1 \geq 4 (abc)^{3/4}$ , thus $abc \geq 1$ .

Now let $t = \sqrt[3]{abc} \geq 1$ , then
$\sum a \geq 3t$ and by Schur,
$\sum ab \leq \frac{\sum a^3 + 6 abc}{\sum a} = \frac{10 abc - 1}{\sum a} \leq \frac{10 t^3 - 1}{3t}$ Substituting the above results and factoring, we obtain
$abc - (2 - a)(2 - b)(2 - c) \geq \frac{2 (t - 1) (3 t^3 - 7 t^2 + 11 t - 1)}{3 t} \geq 0$

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sqing

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#15 h Nov 25, 2024, 7:05 AM

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Very very nice.Thank lbh_qys.

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#16 h Nov 28, 2024, 11:33 AM

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Let $a,b\geq 0$ and $ab( a-b)^2 \geq 1.$ Prove that
$a+b\geq 2$ Let $a,b\geq 0$ and $a^2b^2( a^2+ b^2-ab) \geq\frac{256}{243} .$ Prove that
$a+b\geq 2$

This post has been edited 1 time. Last edited by sqing, Nov 28, 2024, 11:36 AM

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#17 h Nov 28, 2024, 3:00 PM

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sqing wrote:

Let $a,b\geq 0$ and $ab( a-b)^2 \geq 1.$ Prove that
$a+b\geq 2$ Let $a,b\geq 0$ and $a^2b^2( a^2+ b^2-ab) \geq\frac{256}{243} .$ Prove that
$a+b\geq 2$

solution

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#18 h Nov 29, 2024, 12:56 AM

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Very very nice.Thank DAVROS.

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#19 h Nov 29, 2024, 2:14 AM

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sqing wrote:

Let $a,b,c$ be reals such that $a^2+2b^2= 9 ,3b^2+4c^2 = 48$ and $5c^2+6a^2 = 51.$ Prove that $-22\leq a+3b+5c\leq 22$ $-10\leq 3a-2b+c\leq 10$

Solution.First we solve the system below:
$\begin{align*} \begin{cases}a^2+2b^2= 9 \\ 3b^2+4c^2 = 48\\ 5c^2+6a^2 = 51 \end{cases}\iff\begin{cases}a^2=1 \\ b^2= 4\\ c^2= 9 \end{cases}\iff \begin{cases}|a|=1 \\ |b|=2\\ |c|=3. \end{cases} \end{align*}$ Therefore, we get
$\begin{align*} |a+3b+5c|\leq&|a|+3|b|+5|c|=22;\\ |3a-2b+c|\leq& 3|a|+2|b|+|c |=10.\quad \blacksquare \end{align*}$

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#20 h Nov 29, 2024, 2:24 AM

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Very very nice.Thank ytChen.

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#21 h Dec 1, 2024, 2:59 AM

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Let $a,b\geq 0$ and $a+b=2.$ Prove that
$a^3b^3( a^3+ b^3-4ab)\leq\frac{54}{125}$ $a^3b^3( a^3+ b^3-8ab)\leq\frac{54}{343}$ $a^3b^3( a^3+ b^3-9ab)\leq\frac{16}{125}$ $a^3b^3( a^3+ b^3-ab)\leq\frac{432}{343}$

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DAVROS

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#22 h Dec 1, 2024, 6:18 AM

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sqing wrote:

Let $a,b\geq 0$ and $a+b=2.$ Prove that $a^3b^3( a^3+ b^3-4ab)\leq\frac{54}{125}$

solution

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no_room_for_error

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#23 h Dec 1, 2024, 8:58 AM

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sqing wrote:

Let $a,b,c\in [0,1]$ and $a+b+c=2 .$ Prove that
$\dfrac{1}{2}\leq \dfrac{1-a^2}{a+2}+\dfrac{1-b^2}{b+2}+\dfrac{1-c^2}{c+2} \leq \dfrac{5}{8}$

We have $a+b=2-c\geq c$ , so $a,b,c$ are side lengths of a triangle and thus $2(ab+bc+ca)\geq a^2+b^2+c^2$ . Therefore, using Cauchy-Schwarz:

$\begin{aligned} \sum \frac{1-a^2}{a+2} &= \sum \left(\frac{1}{2}-a+\frac{3a}{2(2+a)}\right)\\ &= -\frac{1}{2}+\frac{3}{2}\sum\frac{a}{2a+b+c}\\ &\geq -\frac{1}{2}+\frac{3(a+b+c)^2}{4(a^2+b^2+c^2+ab+bc+ca)}\\ &= -\frac{1}{2}+\frac{3}{2}\left(1-\frac{1}{2+\frac{2(ab+bc+ca)}{a^2+b^2+c^2}}\right)\\ &\geq \frac{1}{2} \end{aligned}$
Right hand side follows from

$\frac{1-a^2}{a+2}\leq -\frac{37}{64}a+\frac{57}{96}$

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#24 h Dec 1, 2024, 10:57 AM

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Very very nice.Thank no_room_for_error.

This post has been edited 2 times. Last edited by sqing, Dec 1, 2024, 10:59 AM

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#25 h Dec 1, 2024, 11:23 AM

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Let $0\leq x\leq \frac{1}{2} .$ Prove that
$\frac{7}{5} >2\sqrt{2x}+3\sqrt{1-\sqrt{2x}}-2\sqrt{2x+1-\sqrt{2x}}\geq 0$ Let $0\leq x\leq \frac{1}{3} .$ Prove that
$\frac{7}{5} >2\sqrt{2x}+3\sqrt{1-\sqrt{2x}}-2\sqrt{2x+1-\sqrt{2x}}\geq 1$

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#26 h Dec 2, 2024, 8:30 AM

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sqing wrote:

Let $0\leq x\leq \frac{1}{2} .$ Prove that $\frac{7}{5} >2\sqrt{2x}+3\sqrt{1-\sqrt{2x}}-2\sqrt{2x+1-\sqrt{2x}}\geq 0$

solution

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#27 h Dec 2, 2024, 10:57 AM

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Very very nice.Thank DAVROS.

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#28 h Dec 2, 2024, 11:18 AM

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Let $a,b \geq 0$ and $a+b+ab=3.$ Prove that
$(a-1)(b+1)(a^2+11 ab+b^2) \leq 18$ $(a-1)(b+1)^2(a^2+8 ab+b^2)\leq 18$ $(a-1)^2(b+1) (a^2+17 ab+b^2) \leq 36$ $(a-1)^2(b+1)^2(a^2+20 ab+b^2) \leq 144$ $a(b+1)(a^2+\frac{80}{9}ab+b^2) \leq 27$ $a(b+1)(a^2+8.9024668 ab+b^2) \leq 27$

This post has been edited 2 times. Last edited by sqing, Dec 2, 2024, 11:34 AM

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#29 h Dec 2, 2024, 11:40 AM

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Let $a,b \geq 0$ and $a+b+ab=3.$ Prove that
$(a+1)^2(b+1)^2(a^2+ 7ab+b^2) \leq 144$ Equality holds when $(a,b)=(0,3 )$ or $(a,b)=(1,1 ).$

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#30 h Dec 2, 2024, 12:33 PM

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Let $a,b \geq 0$ and $a+b=2.$ Prove that
$(a-1)(b-1)^2(a^2+8ab+b^2) \leq 4$ $(a+1)(b+1)^2(a^2+\frac{4}{3}ab+b^2) \leq 36$ $(a+1)(b+1)^2(a^2+2ab+b^2) \leq \frac{1024}{27}$ $(a-1)(b-1)^2(a^2+9ab+b^2) \leq \frac{726}{175}\sqrt{ \frac{33}{35}}$

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no_room_for_error

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#32 h Dec 2, 2024, 12:36 PM

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sqing wrote:

Let $a,b \geq 0$ and $a+b+ab=3.$ Prove that
$(a+1)^2(b+1)^2(a^2+ 7ab+b^2) \leq 144$ Equality holds when $(a,b)=(0,3 )$ or $(a,b)=(1,1 ).$

Just

$a^2+7ab+b^2-9=ab(ab-1)\leq 0$

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#33 h Dec 2, 2024, 12:42 PM

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Nice. Thank no_room_for_error.

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#34 h Dec 2, 2024, 1:15 PM

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Let $a,b \geq 0$ and $a+b+ab=3.$ Prove that
$(a+1)(b+2)(a^2+ \frac{463}{50}ab+b^2) \leq 72$ $(a+1)^2(b+2)^2(a^2+ 11 ab+b^2) \leq 576$ $(a+1)(b+3)(a^2+ 9.9337ab+b^2) \leq 108$ $(a+1)^2(b+3)^2(a^2+ 12 ab+b^2) \leq 1296$

This post has been edited 1 time. Last edited by sqing, Dec 2, 2024, 1:24 PM

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