Inequalities

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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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42601 posts

sqing

#1 h Apr 25, 2025, 9:19 AM

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Let $a,b \in [0 ,1] .$ Prove that
$\frac{a}{ 1-ab+b }+\frac{b }{ 1-ab+a } \leq 2$ $\frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 }+\frac{ab }{2+ab } \leq 1$ $\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+ab }\leq \frac{5}{2}$ $\frac{a}{ 1-ab+b^2 }+\frac{b }{ 1-ab+a^2 }+\frac{1 }{1+2ab }\leq \frac{7}{3}$ $\frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 } +\frac{ab }{1+ab }\leq \frac{7}{6 }$

This post has been edited 4 times. Last edited by sqing, Apr 25, 2025, 9:53 AM

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#2 h Apr 25, 2025, 1:21 PM

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$\frac{a}{ 1-ab+b }+\frac{b }{ 1-ab+a } \leq 2$
Let $c=1-ab$ , so we have to prove $\frac{a}{b+c}+\frac{b}{a+c}\leq 2$ , which leads to $a(a+c)+b(b+c)\leq 2(a+c)(b+c)$ , or after expanding and simplifying:
$a^2+b^2\leq 2ab+ac+bc+c^2$ . Note that $c^2+2ab=1+a^2b^2$ , therefore we have to prove $a^2+b^2\leq 1+a^2b^2 +(a+b)c$ , or
$(a^2-1)(b^2-1)+(a+b)c\geq0$ , which is true since $a,b\in[0,1]$ and $c\geq0$ . Equality holds iff $a=b=1$ .

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sqing

#3 h Apr 25, 2025, 1:56 PM

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Very good.Thanks.

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DAVROS

#4 h Apr 26, 2025, 7:42 AM

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sqing wrote:

Let $a,b \in [0 ,1] .$ Prove that $\frac{a}{ 1+ab+b^2 }+\frac{b }{ 1+ab+a^2 }+\frac{ab }{2+ab } \leq 1$

solution

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sqing

#5 h Apr 26, 2025, 8:14 AM

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Very very nice.Thank DAVROS.

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sqing

#6 h Apr 26, 2025, 12:06 PM

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Let $a,b,c,d\geq -1$ and $a+b+c+d=2.$ Prove that
$a + bc + cd + bcd \leq 6$ $a -ab+bc-cd+ d\leq 9$ $a - bc + cd - bcd \leq \frac{73}{8}$ $a^2+2ab- bc+ 2cd+ d^2\leq\frac{77}{4}$ $a^2+2ab-2bc+2cd+ d^2 \leq 20$ $a^2+2ab-2bc+ cd+ d^2\leq 21$

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DAVROS

#7 h Apr 26, 2025, 3:27 PM

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sqing wrote:

Let $a,b,c,d\geq -1$ and $a+b+c+d=2.$ Prove that $a + bc + cd + bcd \leq 6$

solution

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SineWaveSage2009

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SineWaveSage2009

#8 h Apr 26, 2025, 6:42 PM

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these were fun to solve ngl

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sqing

#9 h Apr 27, 2025, 1:46 AM

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Very very nice.Thank DAVROS.

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sqing

#10 h Apr 27, 2025, 2:17 AM

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Let $a,b\geq 0$ and $a+b=2.$ Prove that
$a^3b^2( 3 a+ b ) \leq \frac{107+51\sqrt{17}}{64}$ $a^3b^2( 2a+b) \leq \frac{2117\sqrt{73}-12881}{1458}$ $a^2b( 3a+ b ) \leq \frac{827+73\sqrt{73}}{256}$ $a^2b( \frac 32a+ b ) \leq \frac{827-73\sqrt{73}}{64}$ Let $a,b\geq 0$ and $a+b+ab=3.$ Prove that
$a^2b ( 2a + b ) \leq 46+13\sqrt{13}-\frac{1}{9} \sqrt{307014+85254\sqrt{13}}$

This post has been edited 2 times. Last edited by sqing, Apr 27, 2025, 3:02 AM

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sqing

#11 h Apr 27, 2025, 12:48 PM

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Let $a,b\geq 0$ and $a^2b( 2a+ b ) \geq 4 .$ Prove that
$a+b\geq 2$ Let $a,b\geq 0$ and $a^2b^2(a^3+ab+b^3) \geq 3 .$ Prove that
$a+b\geq 2$ Let $a,b\geq 0$ and $a^2b^2( 2a^2+b^2)(a^2+2b^2) \geq 9 .$ Prove that
$a+b\geq 2$ Let $a,b\geq 0$ and $ab^2( b +1) \geq 4 .$ Prove that
$a+b+ab\geq 3$ Let $a,b\geq 0$ and $ab( b +1) \geq \frac{9}{4} .$ Prove that
$a+b+ab\geq 3$ Let $a,b\geq 0$ and $a^2b ( a+b^2 ) \geq \frac{76}{27} .$ Prove that
$a+b+ab\geq 3$

This post has been edited 1 time. Last edited by sqing, Apr 27, 2025, 12:51 PM

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DAVROS

#12 h Apr 29, 2025, 1:05 PM

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sqing wrote:

Let $a,b,c,d\geq -1$ and $a+b+c+d=2.$ Prove that $a^2+2ab- bc+ 2cd+ d^2\leq\frac{77}{4}$

solution

This post has been edited 1 time. Last edited by DAVROS, Apr 29, 2025, 1:32 PM

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sqing

#13 h Apr 29, 2025, 1:09 PM

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Very very nice.Thank DAVROS.

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DAVROS

#14 h Apr 29, 2025, 1:30 PM

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sqing wrote:

Let $a,b\geq 0$ and $a+b=2.$ Prove that $a^3b^2( 2a+b) \leq \frac{2117\sqrt{73}-12881}{1458}$

solution

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42601 posts

sqing

#15 h Apr 29, 2025, 1:58 PM

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Very very nice.Thank DAVROS.

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ReticulatedPython

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ReticulatedPython

#16 h Apr 29, 2025, 5:02 PM

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Hi sqing, slightly off topic but where do you get the inequalities from? Do you create them by yourself or do you source them from math competitions?

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212 posts

#17 h Apr 29, 2025, 5:25 PM

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I think he just spends a few hours creating inequalities with the givens lol

although I could be wrong, I think this is the most reasonable possibility since the amount of inequalities that he's posted far exceeds the ones that people have solved and also he posts like five equations per given bound

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