Inequality em613

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Inequality em613

G H J

Reveal topic tags

inequalities proposed

L

G H BBookmark kLocked kLocked NReply

Source: Own

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3428 posts

#1 h Apr 20, 2013, 7:07 AM • 2 Y

Y by Adventure10, Mango247

For $a, b, c, d>0, abcd=1$ prove that
$\frac{1}{a+b+c+3}+\frac{1}{b+c+d+3}+\frac{1}{c+d+a+3}+\frac{1}{d+a+b+3}\le\frac{(a+b+c+d)^2}{24}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2130 posts

Sayan

#2 h Apr 20, 2013, 7:48 AM • 3 Y

Y by Limerent, Adventure10, Mango247

oldbeginner wrote:

For $a, b, c, d>0, abcd=1$ prove that
$\frac{1}{a+b+c+3}+\frac{1}{b+c+d+3}+\frac{1}{c+d+a+3}+\frac{1}{d+a+b+3}\le\frac{(a+b+c+d)^2}{24}$

By AM-GM:
$\sum\frac1{a+b+c+3} \le \frac16\sum\frac1{\sqrt[6]{abc}}=\frac16\sum\sqrt[6]a$
By power mean inequality we have
$\frac{(a+b+c+d)^2}4\ge a+b+c+d \ge \sqrt[6]a+\sqrt[6]b+\sqrt[6]c+\sqrt[6]d$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41050 posts

sqing

#3 h Feb 1, 2020, 3:00 AM • 1 Y

Y by Adventure10

For $a, b, c, d>0, abcd=1.$ Prove that $\frac1{a^4+a^3+ a^2+a}+\frac1{b^4+b^3+b ^2+b}+\frac1{c^4+c^3+c^2+c }+\frac1{d^4+d^3+d ^2+d} \ge 1$ $\frac1{a +1}+\frac1{b+1}+\frac1{c+1 }+\frac1{d+1} \ge 2\left(\frac a { a^4+3}+\frac b { b^4+3}+\frac c{ c^4+3}+\frac d{ d^4+3}\right).$ Old?
Where?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Victoria_Discalceata1

743 posts

Victoria_Discalceata1

#4 h Feb 1, 2020, 4:45 AM • 1 Y

Y by Adventure10

sqing wrote:

Old?
Where?

I would try and look in Vasc's old topics because they both follow by one of his inequalities: if $x_1,x_2,...,x_n>0$ are such that $\prod\limits_{i=1}^nx_i=1$ then $\sum\limits_{i=1}^n\frac{1}{\sum\limits_{k=1}^{n}x_i^{n-k}}\ge 1$ .

We have $\frac{1}{x^4+x^3+x^2+x}\ge\frac{5}{3}\cdot\frac{1}{x^3+x^2+x+1}-\frac{1}{6}$ and $\frac{1}{x+1}-\frac{2x}{x^4+3}\ge\frac{2}{3}\cdot\frac{1}{x^3+x^2+x+1}-\frac{1}{6}$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41050 posts

sqing

#5 h Feb 1, 2020, 5:51 AM • 1 Y

Y by Adventure10

Thank you very much.
We have $\frac{1}{x^4+x^3+x^2+x}\ge\frac{1}{(1+x^\frac{5}{2})^2}$ $\frac1{a^4+a^3+ a^2+a}+\frac1{b^4+b^3+b ^2+b}+\frac1{c^4+c^3+c^2+c }+\frac1{d^4+d^3+d ^2+d}$ $\ge\frac{1}{(1+a^\frac{5}{2})^2}+\frac{1}{(1+b^\frac{5}{2})^2}+\frac{1}{(1+c^\frac{5}{2})^2}+\frac{1}{(1+d^\frac{5}{2})^2}\ge 1$ (Lijvzhi)
For $x_1,x_2,...,x_n>0, \prod\limits_{i=1}^nx_i=1.$ Prove that $\sum\limits_{i=1}^n\frac1{a^4_i+a^3_i+ a^2_i+a_i} \ge \sum\limits_{i=1}^n\frac{1}{(1+a^\frac{5}{2}_i)^2} \ge 1$ h

This post has been edited 3 times. Last edited by sqing, Feb 1, 2020, 2:36 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41050 posts

sqing

#6 h Mar 8, 2020, 8:19 AM

Y by

sqing wrote:

For $a, b, c, d>0, abcd=1.$ Prove that $\frac1{a^4+a^3+ a^2+a}+\frac1{b^4+b^3+b ^2+b}+\frac1{c^4+c^3+c^2+c }+\frac1{d^4+d^3+d ^2+d} \ge 1$

Let $a,b,c,d>0$ such that $a+b+c+d=4$ Prove that
$\frac{1}{a^3+a^2+a+1}+\frac{1}{b^3+b^2+b+1}+\frac{1}{c^3+c^2+c+1}+\frac{1}{d^3+d^2+d+1}\geq1.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41050 posts

sqing

#9 h Jun 15, 2020, 2:55 PM • 1 Y

Y by Mango247

sqing wrote:

For $a, b, c, d>0, abcd=1.$ Prove that $\frac1{a^4+a^3+ a^2+a}+\frac1{b^4+b^3+b ^2+b}+\frac1{c^4+c^3+c^2+c }+\frac1{d^4+d^3+d ^2+d} \ge 1$ $\frac1{a +1}+\frac1{b+1}+\frac1{c+1 }+\frac1{d+1} \ge 2\left(\frac a { a^4+3}+\frac b { b^4+3}+\frac c{ c^4+3}+\frac d{ d^4+3}\right).$ Old?
Where?

For $a, b, c, d>0, abcd=1.$ Prove that $\frac{a}{a^3+3}+\frac{b}{b^3+3}+\frac{c}{c^3+3}+\frac{d}{d^3+3}\le \frac{a}{3a+1}+\frac{b}{3b+1}+\frac{c}{3c+1}+\frac{d}{3d+1}\le1$ $\frac1{a +3}+\frac1{b+3}+\frac1{c+3 }+\frac1{d+3} \leq 1$ SXTX,(4)2020
here

Attachments:

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

39 posts

#10 h Jun 15, 2020, 3:26 PM

Y by

Sayan wrote:

By power mean inequality we have
$\frac{(a+b+c+d)^2}4\ge a+b+c+d \ge \sqrt[6]a+\sqrt[6]b+\sqrt[6]c+\sqrt[6]d$

How did you get $a+b+c+d \ge \sqrt[6]{a}+\sqrt[6]{b}+\sqrt[6]{c}+\sqrt[6]{d}$ ?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

41050 posts

sqing

#11 h Jun 15, 2020, 3:39 PM • 1 Y

Y by Mango247

For $a, b, c, d>0,$ prove that
$\frac{ab}{(s-c)(s-d)}+\frac{ac}{(s-b)(s-d)}+\frac{ad}{(s-b)(s-c)}+\frac{bc}{(s-a)(s-d)}+\frac{bd}{(s-a)(s-c)}+\frac{cd}{(s-a)(s-b)}\leq\frac{2}{3}$ W here $s=a+b+c+d.$
SXTX,(5)2020
ZDSX,(11)2020

Attachments:

This post has been edited 3 times. Last edited by sqing, Jan 3, 2021, 12:11 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

433 posts

NaPrai

#12 h Jun 16, 2020, 2:45 AM

Y by

Limerent wrote:

Sayan wrote:

By power mean inequality we have
$\frac{(a+b+c+d)^2}4\ge a+b+c+d \ge \sqrt[6]a+\sqrt[6]b+\sqrt[6]c+\sqrt[6]d$

How did you get $a+b+c+d \ge \sqrt[6]{a}+\sqrt[6]{b}+\sqrt[6]{c}+\sqrt[6]{d}$ ?

By Power-mean and AM-GM, we have $\sum_{cyc}a \ge \frac{1}{4^5}\left(\sum_{cyc}\sqrt[6]{a}\right)^6 \ge \sum_{cyc}\sqrt[6]{a}$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a