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inequalities
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old and easy imo inequality
Valentin Vornicu   211
N 4 hours ago by Mathdreams
Source: IMO 2000, Problem 2, IMO Shortlist 2000, A1
Let $ a, b, c$ be positive real numbers so that $ abc = 1$. Prove that
\[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1. \]
211 replies
Valentin Vornicu
Oct 24, 2005
Mathdreams
4 hours ago
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Inequalities
lgx57   1
N 5 hours ago by sqing
Let $a,b,c>0$,$\frac{a^2+b^2+c^2}{ab+bc+ca}=2$, find the minimum of

$$\frac{a^3+b^3+c^3}{abc}$$
1 reply
lgx57
5 hours ago
sqing
5 hours ago
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Inequalities
sqing   6
N 5 hours ago 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. $
6 replies
sqing
Yesterday at 3:55 PM
sqing
5 hours ago
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Interesting inequality
sqing   2
N 5 hours ago by sqing
Source: Own
Let $ a,b,c\geq 0,(ab+c^2)(ac+b^2)\neq 0 $ and $ a+b+c=3 . $ Prove that
$$ \frac{1}{ab+c^2}+\frac{1}{ac+b^2} \geq\frac{3}{4} $$$$ \frac{1}{ab+2c^2}+\frac{1}{ac+2b^2} \geq\frac{4}{9} $$
2 replies
sqing
6 hours ago
sqing
5 hours ago
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Elegant inequality
SunnyEvan   4
N 6 hours ago by SunnyEvan
Source: proposed by Zhenping An
Let $a$, $b$, $c$, $d$ be non-negative real numbers such that
\[2a+2b+2c+2d+ab+bc+cd+da+3=abcd.\]prove that : \[\sqrt[4]{abc}+\sqrt[4]{bcd}+\sqrt[4]{cda}+\sqrt[4]{dab}\le\sqrt[4]{27(1+a)(1+b)(1+c)(1+d)}.\]
4 replies
SunnyEvan
Yesterday at 11:32 AM
SunnyEvan
6 hours ago
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Cyclic ine
m4thbl3nd3r   8
N 6 hours ago by m4thbl3nd3r
Let $a,b,c>0$ such that $a+b+c=3$. Prove that $$a^3b+b^3c+c^3a+9abc\le 12$$May everyone not to upload solutions on this problem anymore until May 15/2025, this is an active problem on Mathematical Reflection! (Thank you Victoria_Discalceata1)
8 replies
m4thbl3nd3r
Yesterday at 3:17 PM
m4thbl3nd3r
6 hours ago
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Interesting inequality
sqing   8
N 6 hours ago by SunnyEvan
Source: Own
Let $ a,b,c\geq 0,(ab+c)(ac+b)\neq 0 $ and $ a+b+c=3 . $ Prove that
$$ \frac{1}{ab+kc}+\frac{1}{ac+kb} \geq\frac{4}{3k} $$Where $ k\geq 3. $
$$ \frac{1}{ab+2c}+\frac{1}{ac+2b} \geq\frac{16}{25} $$$$ \frac{1}{ab+3c}+\frac{1}{ac+3b} \geq\frac{4}{9} $$$$ \frac{1}{ab+4c}+\frac{1}{ac+4b} \geq\frac{1}{3} $$

8 replies
sqing
Today at 3:42 AM
SunnyEvan
6 hours ago
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k Olympiad question
slimshady360   4
N Today at 1:24 PM by sqing
Let a,b,c be positive real numbers such that a + b+c = 3abc. Prove that
a2 +b2 +c2 +3 ≥2(ab+bc+ca)
4 replies
slimshady360
Today at 10:28 AM
sqing
Today at 1:24 PM
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Interesting inequality
sqing   2
N Today at 12:58 PM by sqing
Source: Own
Let $ a,b,c\geq 0,(ab+c)(ac+b)\neq 0 $ and $ a+b+c=3 . $ Prove that
$$ \frac{1}{ab+c+k}+\frac{1}{ac+b+k} \geq\frac{2}{k+2} $$Where $ k\geq 0. $
2 replies
sqing
Today at 2:56 AM
sqing
Today at 12:58 PM
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help!!!!!!!!!!!!
Cobedangiu   1
N Today at 12:41 PM by pooh123
help
1 reply
Cobedangiu
Today at 12:32 PM
pooh123
Today at 12:41 PM
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Double factorial inequality
Snoop76   2
N Today at 6:44 AM by Snoop76
Source: own
Show that: $$2n \cdot \sum_{k=0}^n (2k-1)!!{n\choose k}>\sum_{k=0}^n (2k+1)!!{n\choose k}$$Note: consider $(-1)!!=1$ and $n>1$
2 replies
Snoop76
Feb 7, 2025
Snoop76
Today at 6:44 AM
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Inequality with real numbers
JK1603JK   4
N Today at 5:53 AM by SunnyEvan
Source: unknown
Let a,b,c are real numbers. Prove that (a^3+b^3+c^3+3abc)^4+(a+b+c)^3(a+b-c)^3(-a+b+c)^3(a-b+c)^3>=0
4 replies
JK1603JK
Yesterday at 6:48 AM
SunnyEvan
Today at 5:53 AM
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Inequalities
sqing   3
N Mar 18, 2025 by sqing
Let $ a,b\geq 2 . $ Prove that
$$ (1-a^2)(1-b^2) (ab-1)^2-108 ab \geq-351 $$$$(1-a^2)(1-b^2) (ab-1)(a^2b^2-1)- 621 ab \geq-2079 $$
3 replies
sqing
Mar 18, 2025
sqing
Mar 18, 2025
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Inequalities
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inequalities
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sqing
41195 posts
sqing
#1 h  Mar 18, 2025, 9:03 AM
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Let $ a,b\geq 2 . $ Prove that
$$ (1-a^2)(1-b^2) (ab-1)^2-108 ab \geq-351 $$$$(1-a^2)(1-b^2) (ab-1)(a^2b^2-1)- 621 ab \geq-2079 $$
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lbh_qys
425 posts
lbh_qys
#2 h  Mar 18, 2025, 9:15 AM
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sqing wrote:
Let $ a,b\geq 2 . $ Prove that
$$ (1-a^2)(1-b^2) (ab-1)^2-108 ab \geq-351 $$
Let $f(a,b) = (1-a^2)(1-b^2) (ab-1)^2-108 ab$, then
$$ f_a(a,b) = (b^2-1)\left[2a\times (ab-1)^2 + (a^2-1)\times 2b(ab-1)\right] - 108\,b \ge \left(\frac{3}{2}\,b\right)\left(4\cdot9+3\cdot2\cdot2\cdot3\right) - 108\,b = 0 $$
Therefore, the $f(a,b)$ is monotonically increasing with respect to \(a\) and, similarly, with respect to \(b\), which implies that

$$ f(a,b) \ge f(2,2) = -351. $$
This post has been edited 1 time. Last edited by lbh_qys, Mar 18, 2025, 9:16 AM
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sqing
41195 posts
sqing
#3 h  Mar 18, 2025, 9:22 AM
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Very very nice.Thank lbh_qys.
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sqing
41195 posts
sqing
#4 h  Mar 18, 2025, 9:22 AM
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Let $ a,b\geq 2 . $ Prove that
$$ (1-a^3)(1-b^3) (ab-1)^2-672ab \geq-2247 $$$$ (1-a^3)(1-b^3) (ab-1)^3-2457 ab \geq-8505 $$
This post has been edited 1 time. Last edited by sqing, Mar 18, 2025, 9:28 AM
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