Advanced topics in Inequalities

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va2010

1276 posts

va2010

#1 h Mar 7, 2015, 4:43 AM • 11 Y

Y by DrMath, jh235, TheCrafter, Eugenis, mathleticguyyy, somebodyyouusedtoknow, Adventure10, Mango247, NicoN9, cubres, ehuseyinyigit

So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!

Attachments:

advanced-topics-inequalities (8).pdf (139kb)

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tastymath75025

3223 posts

tastymath75025

#2 h Mar 7, 2015, 5:06 AM • 1 Y

Y by Adventure10

Will edit later maybe

4.1: sol

5.1: sol

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This post has been edited 2 times. Last edited by tastymath75025, Mar 7, 2015, 5:31 AM

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Nuran2010

85 posts

Nuran2010

#3 h Apr 16, 2025, 12:30 PM • 1 Y

Y by TunarHasanzade

3.1:Click to reveal hidden text

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Nuran2010

85 posts

Nuran2010

#4 h Apr 16, 2025, 12:43 PM • 2 Y

Y by TunarHasanzade, Strangett

Similar logic,but a bit different approach for 5.1:
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sqing

42102 posts

sqing

#5 h Apr 16, 2025, 1:16 PM • 1 Y

Y by Strangett

Problem 5.1. (Tuan Le)
Let $a,b,c$ be positive real numbers such that $abc\geq 1$ Prove the inequality $\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\le 1.$
Problem4.2. (T.Q.Anh)
Let $a,b,c>0$ and $ab+bc+ca=3$ .Prove that $\frac{a}{2a+b^2}+\frac{b}{2b+c^2}+\frac{c}{2c+a^2} \leq 1$
Problem 3.1. (Titu Andresscu)
Let $a,b,c>0$ and $a+b+c\geq 3 .$ Prove that $\frac{1}{a^2+b+c}+\frac{1}{b^2+c+a}+\frac{1}{c^2+a+b}\leq 1$ https://artofproblemsolving.com/community/c6h1838005p12333824

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giangtruong13

145 posts

giangtruong13

#6 h Apr 16, 2025, 1:48 PM

Y by

You see NOTHING

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Primeniyazidayi

98 posts

Primeniyazidayi

#7 h Apr 16, 2025, 2:54 PM

Y by

The direction of 3.2 is false.Anyway,here is an approach from "Problems from the book,Titu Andreescu and Gabriel Dospinescu"
S3.2

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sqing

42102 posts

sqing

#8 h Apr 16, 2025, 3:14 PM

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If $a,b,c$ are reals show that:
$\frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$

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sqing

42102 posts

sqing

#9 h Apr 16, 2025, 3:23 PM

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Let $x,y,z>0$ and $xyz=x+y+z$ .Prove that $\frac{(x-1)^2}{x^2+1}+\frac{(y-1)^2}{y^2+1}+\frac{(z-1)^2}{z^2+1} \geq 3-\frac{3\sqrt 3}{2}$ *
Let $a, b, c\geq -\frac{3}{4},a+b+c=1 .$ Prove that
$\frac{(a-1)^{2}}{a^2+1 }+\frac{(b-1)^{2}}{b^2+1 }+\frac{(c-1)^{2}}{c^2+1 }\geq \frac{6}{5}$ Let $a, b, c>0,abc\geq 8 .$ Prove that $\dfrac{(a-1)^2}{a^2+2}+\dfrac{(b-1)^2}{b^2+2}+\dfrac{(c-1)^2}{c^2+2}\geq\frac{25}{38}$