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jlacosta   0
May 1, 2025
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0 replies
jlacosta
May 1, 2025
0 replies
Interesting inequalities
sqing   1
N 7 minutes ago by sqing
Source: Own
Let $ a,b,c\geq 0 , (a+k )(b+c)=k+1.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2k-3+2\sqrt{k+1}}{3k-1}$$Where $ k\geq \frac{2}{3}.$
Let $ a,b,c\geq 0 , (a+1)(b+c)=2.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 2\sqrt{2}-1$$Let $ a,b,c\geq 0 , (a+3)(b+c)=4.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{7}{4}$$Let $ a,b,c\geq 0 , (3a+2)(b+c)= 5.$ Prove that
$$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq  \frac{2(2\sqrt{15}-5)}{3}$$
1 reply
1 viewing
sqing
7 minutes ago
sqing
7 minutes ago
Prime sums of pairs
Assassino9931   1
N 11 minutes ago by Nuran2010
Source: Al-Khwarizmi Junior International Olympiad 2025 P5
Sevara writes in red $8$ distinct positive integers and then writes in blue the $28$ sums of each two red numbers. At most how many of the blue numbers can be prime?

Marin Hristov, Bulgaria
1 reply
+1 w
Assassino9931
4 hours ago
Nuran2010
11 minutes ago
Curious inequality
produit   0
12 minutes ago
Positive real numbers x_1, x_2, . . . x_n satisfy x_1 + x_2 + . . . + x_n = 1.
Prove that
1/(1 −√x_1)+1/(1 −√x_2)+ . . . +1/(1 −√x_n)⩾ n + 4.
0 replies
produit
12 minutes ago
0 replies
the number of fractions in lowest terms in (0,1) s.t. fractional part >=1/2
tom-nowy   0
35 minutes ago
Source: Own
Let $n$ be a positive integer. Find the number of reduced fractions $0<p/q<1$ for which $\left\{ n/q \right\} \geq 1/2$, where $\left\{ x \right\}$ denotes the fractional part of $x$. Express your answer in terms of $n$.
0 replies
tom-nowy
35 minutes ago
0 replies
No more topics!
Advanced topics in Inequalities
va2010   23
N Apr 22, 2025 by Novmath
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!
23 replies
va2010
Mar 7, 2015
Novmath
Apr 22, 2025
Advanced topics in Inequalities
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G H BBookmark kLocked kLocked NReply
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va2010
1276 posts
#1 • 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
#2 • 1 Y
Y by Adventure10
Will edit later maybe

4.1: sol

5.1: sol

[/hide]
This post has been edited 2 times. Last edited by tastymath75025, Mar 7, 2015, 5:31 AM
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Nuran2010
84 posts
#3 • 1 Y
Y by TunarHasanzade
3.1:Click to reveal hidden text
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Nuran2010
84 posts
#4 • 2 Y
Y by TunarHasanzade, Strangett
Similar logic,but a bit different approach for 5.1:
Click to reveal hidden text
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sqing
42096 posts
#5 • 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
This post has been edited 2 times. Last edited by sqing, Apr 16, 2025, 2:33 PM
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giangtruong13
145 posts
#6
Y by
You see NOTHING
This post has been edited 1 time. Last edited by giangtruong13, Apr 16, 2025, 1:55 PM
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Primeniyazidayi
98 posts
#7
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
42096 posts
#8
Y by
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
42096 posts
#9
Y by
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 $$\dfrac{(a-1)^2}{a^2+2}+\dfrac{(b-1)^2}{b^2+2}+\dfrac{(c-1)^2}{c^2+2}\geq\frac{25}{38}$$
This post has been edited 1 time. Last edited by sqing, Apr 16, 2025, 3:27 PM
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Strangett
2 posts
#10 • 1 Y
Y by Nuran2010
Problem 3.4. Given that \( a + b + c = 3 \), prove that
\[
(2a^2 + 3)(2b^2 + 3)(2c^2 + 3) \ge 125.
\]
Click to reveal hidden text
This post has been edited 1 time. Last edited by Strangett, Apr 16, 2025, 5:18 PM
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Novmath
20 posts
#12 • 1 Y
Y by Strangett
Strangett wrote:
Problem 3.4. Given that \( a + b + c = 3 \), prove that
\[
(2a^2 + 3)(2b^2 + 3)(2c^2 + 3) \ge 125.
\]
Click to reveal hidden text

Nice solution ,can you explain how did you think of taking f(x)= something with ln. When do we take f(x) as ln(something) ?
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Novmath
20 posts
#13
Y by
va2010 wrote:
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!

The problem 2.1 (USAMO 2001 P3) is wrong. The condition was $a^2+b^2+c^2+ abc =4$
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Strangett
2 posts
#14
Y by
Novmath wrote:
Strangett wrote:
Problem 3.4. Given that \( a + b + c = 3 \), prove that
\[
(2a^2 + 3)(2b^2 + 3)(2c^2 + 3) \ge 125.
\]
Click to reveal hidden text

Nice solution ,can you explain how did you think of taking f(x)= something with ln. When do we take f(x) as ln(something) ?

Just in order to use the sum property of the natural logarithm (to apply Jensen's inequality to sums of function values). Also when taking the derivatives it is pretty obvious we are going to get something with its denominator squared and it is surely a convex function.
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sqing
42096 posts
#15
Y by
Novmath wrote:
va2010 wrote:
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!

The problem 2.1 (USAMO 2001 P3) is wrong. The condition was $a^2+b^2+c^2+ abc =4$
Yeah:
Let $a, b, c \geq 0$ and satisfy $ a^2+b^2+c^2 +abc = 4 . $ Show that\[ 0 \le ab + bc + ca - abc \leq 2. \]
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sqing
42096 posts
#16
Y by
Let $a, b, c \geq 0$ and $ a^2+b^2+c^2 +2abc = 4 . $ Show that$$ 0 \le ab + bc + ca - abc \leq 2$$
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Novmath
20 posts
#17
Y by
Strangett wrote:
Just in order to use the sum property of the natural logarithm (to apply Jensen's inequality to sums of function values). Also when taking the derivatives it is pretty obvious we are going to get something with its denominator squared and it is surely a convex function.
Thank you for explanation.
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Novmath
20 posts
#18
Y by
sqing wrote:
Novmath wrote:
va2010 wrote:
So a while ago, I compiled some tricks on inequalities. You are welcome to post solutions below!

The problem 2.1 (USAMO 2001 P3) is wrong. The condition was $a^2+b^2+c^2+ abc =4$
Yeah:
Let $a, b, c \geq 0$ and satisfy $ a^2+b^2+c^2 +abc = 4 . $ Show that\[ 0 \le ab + bc + ca - abc \leq 2. \]

Among a, b, c there must exist 2 numbers, say a and b, such that $ (1-a)(1-b) \geq 0$
Then: $ 1 \geq a + b - ab$
On the other hand, from $ a^2 + b^2 + c^2 + abc = 4$ we have:
$ c = \frac {-ab + \sqrt{a^2b^2 - 4a^2 - 4b^2 + 16}}{2} \leq \frac {-ab + \sqrt{a^2b^2 - 8ab + 16}}{2} = -ab +2$
Thus:
$ ab + bc + ca - abc = ab + c(a+b-ab) \leq ab + c \leq 2  (q.e.d)$
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Novmath
20 posts
#19
Y by
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
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sqing
42096 posts
#20
Y by
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474
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Novmath
20 posts
#21
Y by
sqing wrote:
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474

Thank you very much :coolspeak:
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Novmath
20 posts
#22
Y by
sqing wrote:
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474

Bro where can I learn separation trick in inequalities ,do you have some source which explaines it detailed, if you send it to me it will be appreciated!
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Novmath
20 posts
#23
Y by
sqing wrote:
Let $a, b, c \geq 0$ and $ a^2+b^2+c^2 +2abc = 4 . $ Show that$$ 0 \le ab + bc + ca - abc \leq 2$$

Can you also share solution of this problem:
(Vasc) Prove that if \( a, b, c \geq 0 \) and
\[
x = a + \frac{1}{b}, \quad y = b + \frac{1}{c}, \quad z = c + \frac{1}{a},
\]then
\[
xy + yz + zx \geq 2 + x + y + z.
\]
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Primeniyazidayi
98 posts
#24
Y by
Solution 3.3
This post has been edited 1 time. Last edited by Primeniyazidayi, Apr 21, 2025, 8:34 PM
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Novmath
20 posts
#25
Y by
sqing wrote:
Novmath wrote:
sqing wrote:
If $ a,b,c$ are reals show that:
$$ \frac{(b+c-a)^2}{(b+c)^2+a^2}+\frac{(c+a-b)^2}{(c+a)^2+b^2}+\frac{(a+b-c)^2}{(a+b)^2+c^2}\geq\frac{3}{5}$$

Can you share solution of this one:
Problem: Prove that for all real numbers \( a, b, c \), the following inequality holds:
\[
4(1 + a^2)(1 + b^2)(1 + c^2) \geq 3(a + b + c)^2
\]
https://artofproblemsolving.com/community/c6h1413227p7963638
https://artofproblemsolving.com/community/c6h1228946p6197808
https://artofproblemsolving.com/community/c6h2239135p17158474

Please share the solution of Vasc one that I shared
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