Inequalities

Art of Problem Solving

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

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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42235 posts

sqing

#1 h Mar 26, 2025, 11:34 AM

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Let $a,b,c\geq 0$ and $a+b+c=1$ . Prove that
$a(b+c+ 5bc +1)\leq\frac{676}{675}$ $a(b+c+6bc +1)\leq\frac{245}{243}$

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42235 posts

sqing

#2 h Mar 26, 2025, 11:58 AM

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Let $a,b,c\geq 0$ and $a(b+c+ 7bc +1)\geq\frac{50}{49}$ . Prove that
$a+b+c \geq 1$

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11352 posts

#3 h Mar 27, 2025, 1:40 AM

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I also had $a,b,c\ge0$ , $a(b+c+2bc+1)\ge1$ implies $a+b+c\ge1$ .

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42235 posts

sqing

#4 h Mar 27, 2025, 2:39 AM

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Thanks.
Let $a,b,c\geq 0$ and $a(b+c +1)\geq 1.$ Prove that
$a+b+c \geq 1$

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11352 posts

#5 h Mar 27, 2025, 3:04 AM

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sqing wrote:

Thanks.
Let $a,b,c\geq 0$ and $a(b+c +1)\geq 1.$ Prove that
$a+b+c \geq 1$

True.

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481 posts

#6 h Mar 27, 2025, 3:29 AM

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sqing wrote:

Thanks.
Let $a,b,c\geq 0$ and $a(b+c +1)\geq 1.$ Prove that
$a+b+c \geq 1$

$b+c \geq \frac{1}{a} - 1$ And so $a+b+c \geq a + \frac{1}{a} -1 \geq 1,$ where the last inequality follows from $a+\frac{1}{a} \geq 2.$

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42235 posts

sqing

#7 h Mar 27, 2025, 11:12 AM

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Nice.Thanks.
$a>0,b+c \geq \frac{1}{a} - 1$ $a+b+c \geq a + \frac{1}{a} -1 \geq 1$

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1693 posts

DAVROS

#8 h Mar 29, 2025, 10:28 AM

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sqing wrote:

Let $a,b,c\geq 0$ and $a(b+c+ 7bc +1)\geq\frac{50}{49}$ . Prove that $a+b+c \geq 1$

solution

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123 posts

#9 h Mar 29, 2025, 11:40 AM

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sqing wrote:

Let $a,b,c\geq 0$ and $a(b+c +1)\geq 1.$ Prove that
$a+b+c \geq 1$

Solution

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42235 posts

sqing

#10 h Mar 29, 2025, 11:49 AM

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

This post has been edited 1 time. Last edited by sqing, Apr 2, 2025, 8:35 AM

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42235 posts

sqing

#11 h Mar 29, 2025, 11:51 AM

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Very very nice.Thank DAVROS.

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42235 posts

sqing

#12 h Mar 29, 2025, 12:25 PM

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Let $a, b, c>0$ and $\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\leq1$ . Prove that
$ab+bc+ca\leq 3$

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