a+b+c=3 inequality

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k a My Retirement & New Leadership at AoPS

rrusczyk 1573

N 6 hours ago by SmartGroot

I write today to announce my retirement as CEO from Art of Problem Solving. When I founded AoPS 22 years ago, I never imagined that we would reach so many students and families, or that we would find so many channels through which we discover, inspire, and train the great problem solvers of the next generation. I am very proud of all we have accomplished and I’m thankful for the many supporters who provided inspiration and encouragement along the way. I'm particularly grateful to all of the wonderful members of the AoPS Community!

I’m delighted to introduce our new leaders - Ben Kornell and Andrew Sutherland. Ben has extensive experience in education and edtech prior to joining AoPS as my successor as CEO, including starting like I did as a classroom teacher. He has a deep understanding of the value of our work because he’s an AoPS parent! Meanwhile, Andrew and I have common roots as founders of education companies; he launched Quizlet at age 15! His journey from founder to MIT to technology and product leader as our Chief Product Officer traces a pathway many of our students will follow in the years to come.

Thank you again for your support for Art of Problem Solving and we look forward to working with millions more wonderful problem solvers in the years to come.

And special thanks to all of the amazing AoPS team members who have helped build AoPS. We’ve come a long way from here:IMAGE

1573 replies

rrusczyk

Mar 24, 2025

SmartGroot

6 hours ago

k a March Highlights and 2025 AoPS Online Class Information

jlacosta 0

Mar 2, 2025

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Do you have plans this summer? There are so many options to fit your schedule and goals whether attending a summer camp or taking online classes, it can be a great break from the routine of the school year. Check out our summer courses at AoPS Online, or if you want a math or language arts class that doesn’t have homework, but is an enriching summer experience, our AoPS Virtual Campus summer camps may be just the ticket! We are expanding our locations for our AoPS Academies across the country with 15 locations so far and new campuses opening in Saratoga CA, Johns Creek GA, and the Upper West Side NY. Check out this page for summer camp information.

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0 replies

jlacosta

Mar 2, 2025

0 replies

SOLVE-3 tasks

Kastriot 2

N an hour ago by sqing

1.Find all pairs of $(a,b)$ which are set of natural numbers and for that $\frac{1}{a}+\frac{1}{b}+\frac{a}{1+b}$ to be a natural number too.

2.Let $a,b,c$ are set of positiv real numbers and $a+b+c=3$ .Then prove $\sqrt{a}+\sqrt{b}+\sqrt{c}\geq ab+bc+ca$ .

3..Let $a,b,c$ are set of positiv real numbers and $a^2+b^2+c^2+(a+b+c)^2\leq4$ .Then prove $\frac{ab+1}{(a+b)^2}+\frac{bc+1}{(b+c)^2}+\frac{ca+1}{(c+a)^2}\geq3$ .

2 replies

Kastriot

Jan 15, 2013

sqing

an hour ago

Inequalities

sqing 5

N 2 hours ago by anduran

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

5 replies

sqing

Yesterday at 11:34 AM

anduran

2 hours ago

Inequalities

sqing 8

N 3 hours ago by sqing

Let $a,b,c\geq 0$ and $a+b+c=1$ . Prove that
$a^2+b^2+ ab +21abc\leq\frac{512}{441}$ Equality holds when $a=b=\frac{38}{21},c=\frac{5}{214}.$
$a^2+b^2+ ab +19abc\leq\frac{10648}{9747}$ Equality holds when $a=b=\frac{22}{57},c=\frac{13}{57}.$
$a^2+b^2+ ab +22abc\leq\frac{15625}{13068}$ Equality holds when $a=b=\frac{25}{66},c=\frac{8}{33}.$

8 replies

sqing

Yesterday at 3:07 AM

sqing

3 hours ago

functions false or true

Math2030 3

N 3 hours ago by jasperE3

find all functions f: \mathbb{R} \to \mathbb{R} that satisfy the functional equation:

f(x^2 f(x) + f(y)) = (f(x))^3 + f(y), \quad \forall x, y \in \mathbb{R}

3 replies

Math2030

Tuesday at 3:05 PM

jasperE3

3 hours ago

An easy inequality

jokehim 3

N Yesterday at 6:53 PM by no_room_for_error

Let $a,b,c\ge 0: a+b+c=3$ then prove that $\color{black}{\frac{a+bc}{b+c+2}+\frac{b+ca}{c+a+2}+\frac{c+ab}{a+b+2}\le \frac{3}{2}.}$

3 replies

jokehim

Yesterday at 1:22 PM

no_room_for_error

Yesterday at 6:53 PM

Inequalities

sqing 13

N Yesterday at 12:18 PM by sqing

Let $a,b,c\geq 0$ and $a+b+c=1$ . Prove that $a^3b+b^3c+c^3a+\frac{473}{256}abc\le\frac{27}{256}$ Equality holds when $a=b=c=\frac{1}{3}$ or $a=0,b=\frac{3}{4},c=\frac{1}{4}$ or $a=\frac{1}{4} ,b=0,c=\frac{3}{4}$
or $a=\frac{3}{4} ,b=\frac{1}{4},c=0.$

13 replies

sqing

Mar 22, 2025

sqing

Yesterday at 12:18 PM

a+b+c=3 inequality

JK1603JK 2

N Yesterday at 3:42 AM by lbh_qys

Let a,b,c\ge 0: a+b+c=3 then prove \frac{a+bc}{b^{2}+c^{2}+2}+\frac{b+ca}{c^{2}+a^{2}+2}+\frac{c+ab}{a^{2}+b^{2}+2}\le \frac{3}{2}
When does equality hold?

2 replies

JK1603JK

Tuesday at 2:04 PM

lbh_qys

Yesterday at 3:42 AM

Ask mininum

TangenT-maTh- 3

N Tuesday at 4:10 PM by rchokler

Find the mininum value of function $f(x)=\cos^2 x-4\cos x-2\sqrt{3}\sin x$

3 replies

TangenT-maTh-

Mar 13, 2025

rchokler

Tuesday at 4:10 PM

Inequalities

sqing 31

N Tuesday at 12:55 PM by sqing

Let $a,b>0$ and $\frac{1}{a}+\frac{1}{b}=1.$ Prove that
$(a^2-a+1)(b^2-b+1) \geq 9$ $(a^2-a+b+1)(b^2-b+a+1) \geq 25$ Let $a,b>0$ and $\frac{1}{a}+\frac{1}{b}=\frac{2}{3}.$ Prove that
$(a+8)(a^2-a+b+2)(b^2-b+5)\geq1331$ $(a+10)(a^2-a+b+4)(b^2-b+7)\geq2197$

31 replies

sqing

Mar 10, 2025

sqing

Tuesday at 12:55 PM

Inequality

MathsII-enjoy 1

N Tuesday at 12:13 PM by sqing

A good inequality problem :coolspeak:

1 reply

MathsII-enjoy

Tuesday at 11:00 AM

sqing

Tuesday at 12:13 PM

an inequality

JK1603JK 1

N Tuesday at 10:18 AM by lbh_qys

Let a,b,c\ge 0: ab+bc+ca>0 then prove \frac{ab+c^2}{a+b}+\frac{bc+a^2}{b+c}+\frac{ca+b^2}{c+a}\ge\frac{2(a^2+b^2+c^2)+ab+bc+ca}{a+b+c}.

1 reply

JK1603JK

Tuesday at 7:56 AM

lbh_qys

Tuesday at 10:18 AM

Inequalities

lgx57 2

N Mar 25, 2025 by zoinkers

Let $a,b,c>0$ , $\frac{a^2+b^2+c^2}{ab+bc+ca}=2$ , find the minimum of

$\frac{a^3+b^3+c^3}{abc}$

2 replies

lgx57

Mar 23, 2025

zoinkers

Mar 25, 2025

inequalities

LuvThoConBoiRoi 12

N Mar 25, 2025 by sqing

With $a,b,c$ are all real positive number that: $abc=a+b+c+2$ . Prove that:
$\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \leq \frac{3}{4}$

12 replies

LuvThoConBoiRoi

Aug 1, 2019

sqing

Mar 25, 2025

a+b+c=3 ine

jokehim 5

N Mar 24, 2025 by LearnMath_105

Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that $\color{black}{\frac{a\left(b+c\right)}{bc+3}+\frac{b\left(c+a\right)}{ca+3}+\frac{c\left(a+b\right)}{ab+3}\le \frac{3}{2}.}$ Proposed by Phan Ngoc Chau

5 replies

jokehim

Mar 18, 2025

LearnMath_105

Mar 24, 2025

a+b+c=3 inequality

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#1 h Tuesday at 2:04 PM

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Let a,b,c\ge 0: a+b+c=3 then prove \frac{a+bc}{b^{2}+c^{2}+2}+\frac{b+ca}{c^{2}+a^{2}+2}+\frac{c+ab}{a^{2}+b^{2}+2}\le \frac{3}{2}
When does equality hold?

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81 posts

#2 h Tuesday at 2:57 PM

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JK1603JK wrote:

Let a,b,c\ge 0: a+b+c=3 then prove \frac{a+bc}{b^{2}+c^{2}+2}+\frac{b+ca}{c^{2}+a^{2}+2}+\frac{c+ab}{a^{2}+b^{2}+2}\le \frac{3}{2}
When does equality hold?

Let $a,b,c\ge 0$ and $a+b+c=3$ . Prove that $\frac{a+bc}{b^{2}+c^{2}+2}+\frac{b+ca}{c^{2}+a^{2}+2}+\frac{c+ab}{a^{2}+b^{2}+2}\le \frac{3}{2}$ When does equality hold?

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438 posts

#3 h Yesterday at 3:42 AM • 1 Y

Y by MihaiT

Let
$A = b^2 + c^2 + 2,\quad B = c^2 + a^2 + 2,\quad C = a^2 + b^2 + 2.$ Then
$9 - 6\sum \frac{a + bc}{A} = \sum\left(3a - \frac{6a + 2(a+b+c)bc}{A}\right) = \sum\frac{a(b-c)^2 + 2b^2(a-c) + 2c^2(a-b)}{A}$ $\geq 2\sum \frac{b^2(a-c)}{A} + 2\sum \frac{c^2(a-b)}{A} = 2\sum \frac{b^2(a-c)}{A} + 2\sum \frac{b^2(c-a)}{C} = 2\sum \frac{b^2(a-c)^2(a+c)}{AC} \geq 0.$
Hence, the original inequality holds, with equality occurring at $(a,b,c)=(1,1,1)$ or $(3,0,0)$ and its cyclic permutations.

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