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inequality
SunnyEvan   9
N 4 minutes ago by SunnyEvan
Let $ x,y \geq 0 ,$ such that : $ \frac{x^2}{x^3+y}+\frac{y^2}{x+y^3} \geq 1 .$
Prove that : $$ x^2+y^2-xy \leq x+y $$$$ (x+\frac{1}{2})^2+(x+\frac{1}{2})^2 \leq \frac{5}{2} $$$$ (x+1)^2+(y+1)^2 \leq 5 $$$$ (x+2)^2+(y+2)^2 \leq 13 $$
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SunnyEvan
Yesterday at 1:51 PM
SunnyEvan
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IMO 2014 Problem 1
Amir Hossein   135
N 8 minutes ago by Kempu33334
Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
Proposed by Gerhard Wöginger, Austria.
135 replies
Amir Hossein
Jul 8, 2014
Kempu33334
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Normal but good inequality
giangtruong13   4
N Apr 23, 2025 by IceyCold
Source: From a province
Let $a,b,c> 0$ satisfy that $a+b+c=3abc$. Prove that: $$\sum_{cyc} \frac{ab}{3c+ab+abc} \geq \frac{3}{5} $$
4 replies
giangtruong13
Mar 31, 2025
IceyCold
Apr 23, 2025
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Normal but good inequality
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Source: From a province
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giangtruong13
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giangtruong13
#1 h  Mar 31, 2025, 4:04 PM
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Let $a,b,c> 0$ satisfy that $a+b+c=3abc$. Prove that: $$\sum_{cyc} \frac{ab}{3c+ab+abc} \geq \frac{3}{5} $$
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arqady
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arqady
#2 h  Apr 1, 2025, 4:11 AM
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giangtruong13 wrote:
Let $a,b,c> 0$ satisfy that $a+b+c=3abc$. Prove that: $$\sum_{cyc} \frac{ab}{3c+ab+abc} \geq \frac{3}{5} $$
After using Bergstrom we obtain something trivial.
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giangtruong13
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giangtruong13
#3 h  Apr 7, 2025, 2:36 PM
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Anyone here?
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Sadigly
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Sadigly
#4 h  Apr 7, 2025, 3:32 PM • 1 Y
Y by MuradSafarli
$\sum\frac{1}{ab}=3$. Let $x=\frac1{a}$. Define $y,z$ similarly. We got $xy+yz+zx=3$. Problem asks us to prove $$\sum\frac{z}{3xy+z+1}\geq\frac35$$
$3=xy+yz+zx\stackrel{\text{AM-GM}}{\geq}3\sqrt{x^2y^2z^2}\Rightarrow xyz\leq1$

$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)\geq3(xy+yz+zx)\Rightarrow x+y+z\geq3$

$(x+y+z)(x+y+z)\stackrel{\text{C-S}}{\leq} 3(x^2+y^2+z^2)$

$$\sum\frac{z}{3xy+z+1}=\sum\frac{z^2}{3xyz+z^2+z}\stackrel{\text{Titu}}{\geq}\frac{(x+y+z)^2}{x^2+y^2+z^2+x+y+z+9xyz}=\frac{x^2+y^2+z^2+6}{x^2+y^2+z^2+x+y+z+9xyz}=1-\frac{x+y+z+9xyz-6}{x^2+y^2+z^2+x+y+z+9xyz}\stackrel{?}{\geq}\frac35$$
$$\frac{x+y+z+9xyz-6}{x^2+y^2+z^2+x+y+z+9xyz}\stackrel{?}{\leq}\frac25$$
Cross multiplying gives $$2x^2+2y^2+2z^2-3x-3y-3z-27xyz\stackrel{?}{\geq}-30$$$$2x^2+2y^2+2z^2-3x-3y-3z-27xyz\geq 2x^2+2y^2+2z^2-3x-3y-3z-27\stackrel{?}{\geq}-30\Rightarrow 2x^2+2y^2+2z^2-3x-3y-3z\stackrel{?}{\geq}-3$$$$6(x^2+y^2+z^2)-9(x+y+z)\stackrel{?}{\geq}-9$$
$$6(x^2+y^2+z^2)-9(x+y+z)\geq 2(x+y+z)(x+y+z)-9(x+y+z)=(x+y+z)(2x+2y+2z-9)\stackrel{?}{\geq}-9$$
Let $u=x+y+z$. This equation becomes $2u^2-9u+9\stackrel{?}{\geq}0$ or $(u-3)(2u-3)\stackrel{?}{\geq}0$. But this is obviously true since $u=x+y+z\geq3$
This post has been edited 1 time. Last edited by Sadigly, Apr 7, 2025, 3:36 PM
Reason: Sometimes I look in a mirror and ask myself: Am I really scared of passing away? If it's today, I hope I hear a Cry out from heaven so loud it can water down a demon With the holy ghost 'til it drown in the blood of Jesus
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IceyCold
212 posts
IceyCold
#5 h  Apr 23, 2025, 3:56 PM
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giangtruong13 wrote:
Let $a,b,c> 0$ satisfy that $a+b+c=3abc$. Prove that: $$\sum_{cyc} \frac{ab}{3c+ab+abc} \geq \frac{3}{5} $$

From Ha Noi last year?
The inequality seems nice,somewhat easy for said test,but it can be generalized.
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