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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Yesterday at 2:14 PM
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

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MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Yesterday at 2:14 PM
0 replies
I’m desperate, Can someone pls help me out?
Dakrya   2
N 6 minutes ago by alexheinis
Let A = {1,5,7,{1},{7},{1,5},{5,7}}
Let G = {B⊂P(A)| n(A-B) = 4} when P(A) is a power set of set A. Find n(G)
2 replies
Dakrya
Jun 27, 2025
alexheinis
6 minutes ago
Challenge: Make as many positive integers from 2 zeros
Biglion   7
N 14 minutes ago by Ravensrule8
How many positive integers can you make from at most 2 zeros, any math operation and cocatination?

Starting from 0, 0=0
7 replies
Biglion
Today at 4:48 PM
Ravensrule8
14 minutes ago
Trigonometry
Maaaaaaath   4
N 3 hours ago by vanstraelen
In a geometric progression we have $ a_1 = \sin x $ , $a_2 = \cos x $ and $a_3 = \tan x$ .Which term of this progression is equal to $1+\cos x$ ?
4 replies
Maaaaaaath
5 hours ago
vanstraelen
3 hours ago
inequality
adamath4ever   2
N 4 hours ago by adamath4ever
$0<a,b,c<1$ real numbers ; such that : $2(a+b+c)+4abc = 3(ab +ac+bc)+1$
Prove that: $a+b+c<3/4$
2 replies
adamath4ever
Today at 5:09 PM
adamath4ever
4 hours ago
Trigonometry
Maaaaaaath   0
5 hours ago
In a geometric progression we have $ a_1 = \sin x $ , $a_2 = \cos x $ and $a_3 = \tan x$ .Which term of this progression is equal to $1+\cos x$ ?
0 replies
Maaaaaaath
5 hours ago
0 replies
2016 preRMO p9, 4 b^2 -a^3 where a,b roots of x^2 + x - 3 = 0
parmenides51   9
N 6 hours ago by fathersayno
Let $a$ and $b$ be the roots of the equation $x^2 + x - 3 = 0$. Find the value of the expression $4b^2 -a^3$.
9 replies
parmenides51
Aug 9, 2019
fathersayno
6 hours ago
random thing I found
sv_legend   1
N Today at 5:27 PM by nudinhtien
This is probably useless but I found that:

If you have 2 squares $n^2$ and $(n+k)^2$, the distance between them is $2(n+1)-1 + 2(n+2)-1 + 2(n+3)-1+...+2(n+k-1)-1+2(n+k)-1$ because I know the distance between $(n-1)^2$ and $n^2$ is $2n-1$ or in other words the $n$th odd number so I'm just adding all of the distances between consecutive squares. (wait can't you just use difference of squares) Yeah so this is just a way longer way to find distance between squares rather than just using difference of squares and factoring.


I hope this is actually true otherwise I will lose so much aura
1 reply
sv_legend
Today at 4:30 PM
nudinhtien
Today at 5:27 PM
Find line EF
Darealzolt   0
Today at 5:17 PM
4 Points \(A,B,C,D\) lie on a circle in that order shown, the extension of \(AB\) meets the extension of \(DC\) at \(E\), the extension of of \(AD\) meets the extension of \(BC\) at \(F\). Let \(EP\) and \(FQ\) be tangents with \(P\) and \(Q\) being the tangent points. If \(EP=60\), and \(FQ = 63\), find the measure of \(EF\)
0 replies
Darealzolt
Today at 5:17 PM
0 replies
Inequalities
sqing   1
N Today at 4:06 PM by DAVROS
Suppose that $x$ and $y$ are nonzero real numbers such that $\left(x + \frac{1}{y} \right) \left(y + \frac{1}{x} \right) = 5$. Prove that
$$\left( x^2-y^2 + \frac{1}{y^2} \right) \left(y^2-x^2 + \frac{1}{x^2} \right)\leq  \frac{7+3\sqrt{5}}{2}$$$$\left( x^3 -y^3+ \frac{1}{y^3} \right) \left(y^3 -x^3+ \frac{1}{x^3} \right)\leq  9+4\sqrt{5}$$
1 reply
sqing
Today at 11:54 AM
DAVROS
Today at 4:06 PM
[PMO25 Qualifying I.8] Length of Diagonal
kae_3   1
N Today at 4:05 PM by vanstraelen
The sides of a convex quadrilateral have lengths $12$cm, $12$cm, $16$cm, and $16$cm, and they are arranged so that there are no pairs of parallel sides. If one of the diagonals is $20$cm long, and the length of the other diagonal is a rational number, what is the length of the other diagonal?

$\text{(a) }\frac{48}{5}\text{ cm}\qquad\text{(b) }\frac{84}{5}\text{ cm}\qquad\text{(c) }\frac{96}{5}\text{ cm}\qquad\text{(d) }\frac{108}{5}\text{ cm}$

Answer Confirmation
1 reply
kae_3
Feb 23, 2025
vanstraelen
Today at 4:05 PM
Root of Unity filtering
Kyj9981   5
N Today at 4:02 PM by Andyluo
Evaluate
$\binom{2017}{0}+\binom{2017}{3}+\cdots+\binom{2017}{2016}$

Source: Ray Li Handout on Roots of Unity

Answer
5 replies
Kyj9981
Today at 3:22 PM
Andyluo
Today at 4:02 PM
Original Problem # 2
ondynarilyChezy   1
N Today at 3:43 PM by ondynarilyChezy
How many ways can you express the number $28$ as the sum of $4$ even positive integers?
1 reply
ondynarilyChezy
Today at 3:35 PM
ondynarilyChezy
Today at 3:43 PM
Average divisors
Magdalo   1
N Today at 3:38 PM by Magdalo
I have a number $n$, where the the greatest common divisor of $10!$ and $n$ is equal to the least common multiple of $5!$ and $n$. What is the average amount of divisors of $n$?
1 reply
Magdalo
Today at 3:36 PM
Magdalo
Today at 3:38 PM
Penchick Porridge
Magdalo   1
N Today at 3:35 PM by Magdalo
Penrick has $9$ cans of porridge, with $3$ each of red, blue, and green cans. He places them randomly in a $3\times3$ grid. What is the expected sum of distinct colors per column? For example, an arrangement with $3$ of the same color per column has a sum of $3$.
1 reply
Magdalo
Today at 3:35 PM
Magdalo
Today at 3:35 PM
Inequalities
sqing   20
N Jun 21, 2025 by sqing
Let $ a,b\geq  0,a+b+ab=3 . $ Prove that
$$  \frac{a}{b+1}+\frac{b}{a^2+1}+\frac{1}{a+b^2} \geq \frac{3}{2} $$$$  \frac{10}{3}\geq  \frac{a}{b+1}+\frac{b}{a^2+1}+\frac{1}{a+b^2}+ab \geq \frac{5}{2} $$$$ \frac{a}{ab+1}+\frac{b }{b+1}+\frac{1 }{a+1} \geq \frac{3}{2} $$$$ \frac{13}{4} \geq\frac{a}{ab+1}+\frac{b }{b+1}+\frac{1 }{a+1}+ ab\geq \frac{7}{4} $$
20 replies
sqing
Jun 14, 2025
sqing
Jun 21, 2025
Inequalities
G H J
G H BBookmark kLocked kLocked NReply
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sqing
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#1
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Let $ a,b\geq  0,a+b+ab=3 . $ Prove that
$$  \frac{a}{b+1}+\frac{b}{a^2+1}+\frac{1}{a+b^2} \geq \frac{3}{2} $$$$  \frac{10}{3}\geq  \frac{a}{b+1}+\frac{b}{a^2+1}+\frac{1}{a+b^2}+ab \geq \frac{5}{2} $$$$ \frac{a}{ab+1}+\frac{b }{b+1}+\frac{1 }{a+1} \geq \frac{3}{2} $$$$ \frac{13}{4} \geq\frac{a}{ab+1}+\frac{b }{b+1}+\frac{1 }{a+1}+ ab\geq \frac{7}{4} $$
This post has been edited 3 times. Last edited by sqing, Jun 14, 2025, 9:10 AM
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sqing
43003 posts
#2
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Let $ a,b\geq  0,a+b+ab=3 . $ Prove that
$$ \frac{a^2+b}{a+b}+\frac{b^2+1}{b+1} \geq 2 $$$$4\geq  \frac{a^2+b}{a+b}+\frac{b^2+1}{b+1}+ab  \geq 3 $$
This post has been edited 1 time. Last edited by sqing, Jun 14, 2025, 8:14 AM
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pooh123
128 posts
#3
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sqing wrote:
Let $ a,b\geq  0,a+b+ab=3 . $ Prove that
$$  \frac{a}{b+1}+\frac{b}{a^2+1}+\frac{1}{a+b^2} \geq \frac{3}{2} $$

From the assumption we get \(2 \leq a+b \leq 3\). Let us prove a more general problem:
Let \(a,b,c \geq 0, bc+ca+ab=3, 3 \leq a+b+c \leq 4\), prove that:
\[
\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2} \geq \frac{3}{2}
\]Let \(x=a+b+c \in [3;4]\). Applying the AM-GM and Cauchy-Schwarz inequalities, we have:
\[
\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}=\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2} \geq \frac{(a+b+c)^2}{ab+ac^2+bc+ba^2+ca+cb^2}
\]Since \(bc+ca+ab=3\) and \(ac^2+ba^2+cb^2 \leq \frac{1}{9}(a+b+c)^3\), we just need to prove:
\[
\frac{9x^2}{x^3+27} \geq \frac{3}{2} \Longleftrightarrow 18x^3 \geq 3x^3+81 \Longleftrightarrow (x-3)(9+3x-x^2) \geq 0
\]Obviously the last inequality is true because \(3 \leq x \leq 4\). Equality at \(a=b=c=1\).
This post has been edited 2 times. Last edited by pooh123, Jun 14, 2025, 8:35 AM
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sqing
43003 posts
#4
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Very very nice.Thank pooh123.
Z K Y
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pooh123
128 posts
#5
Y by
sqing wrote:
Let $ a,b\geq  0,a+b+ab=3 . $ Prove that
$$ \frac{a^2+b}{a+b}+\frac{b^2+1}{b+1} \geq 2 $$

For \(a,b,c \geq 0,bc+ca+ab=3\), we'll prove that:
\[
\frac{a^2+b}{a+b}+\frac{b^2+c}{b+c}+\frac{c^2+a}{c+a} \geq 3
\]Let \(x=a+b+c \geq 3\) and applying the Cauchy-Schwarz inequality, we have:
\[
\frac{a^2+b}{a+b}+\frac{b^2+c}{b+c}+\frac{c^2+a}{c+a}=\frac{(a^2+b)(1+b)}{(a+b)(1+b)}+\frac{(b^2+c)(1+c)}{(b+c)(1+c)}+\frac{(c^2+a)(1+a)}{(c+a)(1+a)}
\]\[
\geq \frac{(a+b)^2}{(1+b)(a+b)}+\frac{(b+c)^2}{(1+c)(c+a)}+\frac{(c+a)^2}{(1+a)(a+b)} \geq \frac{4(a+b+c)^2}{a^2+b^2+c^2+bc+ca+ab+2a+2b+2c}
\]Since \(a^2+b^2+c^2+bc+ca+ab=(a+b+c)^2-bc-ca-ab=(a+b+c)^2-3\), we just need to prove:
\[
\frac{4x^2}{x^2+2x-3} \geq 3 \Longleftrightarrow 4x^2 \geq 3x^2+6x-9 \Longleftrightarrow (x-3)^2 \geq 0
\]Obviously the last inequality is true. Equality at \(a=b=c=1\).
This post has been edited 1 time. Last edited by pooh123, Jun 14, 2025, 12:54 PM
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sqing
43003 posts
#6
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Very nice.Thank pooh123:
Let $ a,b,c\geq  0,ab+bc+ca =3 . $ Prove that
$$\frac{a^2+b}{a+b}+\frac{b^2+c}{b+c}+\frac{c^2+a}{c+a} \geq 3$$Let $ a,b,c\geq  0,a+b+c=3 . $ Prove that
$$\frac{a^2+b}{a+b}+\frac{b^2+c}{b+c}+\frac{c^2+a}{c+a} \geq 3$$
This post has been edited 1 time. Last edited by sqing, Jun 15, 2025, 6:44 AM
Z K Y
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sqing
43003 posts
#7
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Let $ a,b,c>0. $ Prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{3}{2}\sqrt{\frac{2c}{a+b}+1}\geq\frac{3}{2}\sqrt[3]{9}$$$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{3}{4}\sqrt{\frac{2c}{a+b}+1}\geq \frac{3}{2}\sqrt[3]{\frac{9}{4}}$$$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{1}{4}\sqrt{\frac{2c}{a+b}+1}\geq\frac{3}{2\sqrt[3]{4}}$$$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{5}{2}\sqrt{\frac{2c}{a+b}+1}\geq\frac{3}{2}\sqrt[3]{25}$$
This post has been edited 1 time. Last edited by sqing, Jun 15, 2025, 9:31 AM
Z K Y
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sqing
43003 posts
#8
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Let $ a,b,c>0. $ Prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+2\sqrt{\frac{3c}{a+b}+\frac{3}{2}}\geq3\sqrt[3]{3}$$$$\frac{a}{b+c}+\frac{b}{c+a}+3\sqrt{\frac{3c}{a+b}+\frac{3}{2}}\geq\frac{9}{\sqrt[3]{4}}$$$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{1}{2}\sqrt{\frac{3c}{a+b}+\frac{3}{2}}\geq \frac{3}{2}\sqrt[3]{\frac{3}{2}}$$$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{1}{3}\sqrt{\frac{3c}{a+b}+\frac{3}{2}}\geq\sqrt[3]{\frac{9}{4}}$$
This post has been edited 1 time. Last edited by sqing, Jun 15, 2025, 8:14 AM
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DAVROS
1747 posts
#9
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sqing wrote:
Let $ a,b,c>0. $ Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{3}{2}\sqrt{\frac{2c}{a+b}+1}\geq\frac{3}{2}\sqrt[3]{4}$
solution
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sqing
43003 posts
#10
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Very very nice.Thank DAVROS.Sorry.
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#11
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Let $ a,b>0,a+b-ab=1. $ Prove that
$$\dfrac{10}{3} \geq \dfrac{a}{b+1}+\dfrac{b}{a+1}+ab-\sqrt{2-ab} \geq 1-\sqrt 2$$Let $ a,b>0,a+b+ab=3. $ Prove that
$$3-\sqrt 2 \geq \dfrac{a}{b+1}+\dfrac{b}{a+1}+ab-\sqrt{2-ab} \geq 1$$
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pooh123
128 posts
#12
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sqing wrote:
Let $ a,b>0,a+b-ab=1. $ Prove that
$$\dfrac{10}{3} \geq \dfrac{a}{b+1}+\dfrac{b}{a+1}+ab-\sqrt{2-ab} \geq 1-\sqrt 2$$

From \(a+b-ab=1\), we get \((a-1)(b-1)=0\) so one of \(a,b\) is equal to \(1\).
WLOG, let \(b=1\), then \(0 \leq a \leq 2\) and the inequality reduces to proving:
\[
\frac{10}{3} \geq \frac{3a}{2}+\frac{1}{a+1}-\sqrt{2-a} \geq 1-\sqrt{2}
\]Minimum: We have:
\[
\frac{3a}{2}+\frac{1}{a+1}-\sqrt{2-a} \geq 1-\sqrt{2} \Longleftrightarrow \frac{3a}{2}+\frac{1}{a+1}-1 \geq \sqrt{2-a}-\sqrt{2}
\]\[
\Longleftrightarrow \frac{a(3a+1)}{2(a+1)} \geq \frac{-a}{\sqrt{2-a}+\sqrt{2}} \Longleftrightarrow \frac{a(3a+1)}{2(a+1)}+\frac{a}{\sqrt{2-a}+\sqrt{2}} \geq 0
\]The last inequality is true because \(0 \leq a \leq 2\). Equality at \(a=0,b=1\).
Maximum: We have:
\[
\frac{3a}{2}+\frac{1}{a+1}-\sqrt{2-a} \leq \frac{10}{3} \Longleftrightarrow \sqrt{2-a} \geq \frac{3a}{2}+\frac{1}{a+1}-\frac{10}{3}
\]\[
\Longleftrightarrow \sqrt{2-a} \geq \frac{(a-2)(9a+7)}{6(a+1)} \Longleftrightarrow \sqrt{2-a}+\frac{(2-a)(9a+7)}{6(a+1)} \geq 0
\]The last inequality is true because \(0 \leq a \leq 2\). Equality at \(a=2,b=1\).
This post has been edited 1 time. Last edited by pooh123, Jun 16, 2025, 1:26 AM
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sqing
43003 posts
#13
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Very very nice.Thank pooh123.
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43003 posts
#14
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Let $ a, b,c> 0, (a+b+c) \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=10 . $ Prove that
$$  \frac{17}{16} \geq \frac{ a^5+b^5+c^5}{abc(a+b+c)^2} \geq   \frac{13}{20}$$$$   \frac{ a^5+b^5+c^5}{(ab+bc+ca)(a+b+c)^3} \geq   \frac{13}{200}$$$$ \frac{ a^5+b^5+c^5}{(a^2+b^2+c^2)(a+b+c)^3} \geq   \frac{13}{225}$$
This post has been edited 2 times. Last edited by sqing, Jun 18, 2025, 7:55 AM
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#15
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Let $ a, b,c> 0, (a+b+c) \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=10 . $ Prove that
$$ \frac{971+21\sqrt{21}}{2000}\geq\frac{\frac{3}{2}a^2+ b^2+\frac{3}{2}c^2}{(a+b+c)^2}\geq   \frac{7}{16}$$$$  \frac{9}{25}>  \frac{\frac{12}{5}a^3+ b^3+\frac{12}{5}c^3}{(a+b+c)^3}\geq   \frac{1}{5}$$
This post has been edited 1 time. Last edited by sqing, Jun 18, 2025, 7:59 AM
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DAVROS
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#16
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sqing wrote:
Let $ a, b,c> 0, (a+b+c) \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=10 . $ Prove that $ \frac{ a^5+b^5+c^5}{abc(a+b+c)^2} \geq   \frac{13}{20}$
solution
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#17
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Very very nice.Thank DAVROS:
Let $ a, b,c> 0, (a+b+c) \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=10 . $ Prove that $$ \frac{17}{16} \geq  \frac{ a^5+b^5+c^5}{abc(a+b+c)^2} \geq   \frac{13}{20}$$
This post has been edited 1 time. Last edited by sqing, Jun 18, 2025, 7:07 AM
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DAVROS
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#18
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sqing wrote:
Let $ a, b,c> 0, (a+b+c) \left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)=10 . $ Prove that $ \frac{ a^5+b^5+c^5}{(a^2+b^2+c^2)(a+b+c)^3} \geq   \frac{13}{225}$
solution
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Very very nice.Thank DAVROS.
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#20
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Let $  a,b,c \ge 0,  a^2 + b^2 + c^2=1 $. Prove that
$$   \frac {1}{2 - a(b+c)} + \frac {1}{2 - b(c+a)} + \frac {1}{2 - c(a+b)} \leq \frac{9}{4}$$
This post has been edited 1 time. Last edited by sqing, Jun 19, 2025, 2:09 PM
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#21
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Let $ a,b,c> 0 .$ Prove that$$ \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{a^2+b^2+2c(a+b)}{ab+bc+ ca}+1$$$$ \frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ca}{b^2}\geq \frac{a^2+b^2+2c(a+b)}{ab+bc+ ca}+1$$
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