I wanna help

by AlastorMoody, Dec 3, 2019, 5:23 AM

I came across a difficulty faced by many people. There's no collection of problems in a particular topic that occured in some particular year. This is my attempt at collecting SOME inequalities from the year 2018-2019

CWMI 2019 P4 wrote:

Let $n$ be a given integer such that $n\ge 2$ . Find the smallest real number $\lambda$ with the following property: for any real numbers $x_1,x_2,\ldots ,x_n\in [0,1]$ , there exists integers $\varepsilon_1,\varepsilon_2,\ldots ,\varepsilon_n\in\{0,1\}$ such that the inequality $\left\vert \sum^j_{k=i} (\varepsilon_k-x_k)\right\vert\le \lambda$ holds for all pairs of integers $(i,j)$ where $1\le i\le j\le n$ .
Bulgaria JBMO TST 2019 wrote:

Let $a,b,c\geq 1.$ Find the maximum value of $\frac{\sqrt{a-1}}{a+b}+\frac{\sqrt{b-1}}{b+c}+\frac{\sqrt{c-1}}{c+a}.$
ARMO 2019 Grade 9 P6 wrote:

For $a,b,c$ real numbers not less than $1$ , prove that $\frac{a+b+c}{4} \geq \frac{\sqrt{ab-1}}{b+c}+\frac{\sqrt{bc-1}}{c+a}+\frac{\sqrt{ca-1}}{a+b}$ .
KBO 2019 Grade 8 wrote:

Given real numbers $a,b,c$ such that $a+b+c+2=abc$ . Prove that :
$(a^2+1)(b^2+1)(c^2+1) \ge 4$
Spain MO 2019 wrote:

Let $a, b, c> 0 .$ Prove that $\frac{a^2}{b^3c}-\frac{a}{b^2}\geq\frac{c}{b}-\frac{c^2}{a}.$ When does the equality hold ?
Kazakhstan MO 2019 Grade 9 P3 wrote:

Let $a,b,c>0$ and $a+b+c=3$ . Prove the inequality,
$\sqrt[3]{{\frac{1}{{3{a^2}(8b + 1)}}}} + \sqrt[3]{{\frac{1}{{3{b^2}(8c + 1)}}}} + \sqrt[3]{{\frac{1}{{3{c^2}(8a + 1)}}}} \ge 1$ .
Ukraine MO 2019 wrote:

Let $x,y,z$ be positive reals such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3.$ Prove that
$(x-1)(y-1)(z-1)\le\frac{1}{4}(xyz-1).$
Moldova MO 2019 wrote:

For $a,b,c$ - real numbers and
$a+b+c = 4$ $a^2+b^2+c^2 = 6$ Prove that
$\frac{86}{9} \le a^3+b^3+c^3 \le 10$
St. Petersburg MO 2019 wrote:

Let $a, b$ and $c$ be non-zero natural numbers such that $c \geq b$ . Show that
$a^b\left(a+b\right)^c>c^b a^c.$
Peking University Comprehensive Summer Camp 2019 P4 wrote:

$a,b,c$ are three real numbers satisfying $a+b+c$ $=$ $a^2+b^2+c^2=2$ . Find the minimum and maximum of $abc$ .
Root Cup Math Competition 2019 wrote:

Let $a_1,a_2,\cdots,a_n\geq -1 (n\ge 2)$ and $a_1+a_2+\cdots+a_n\geq 0.$ .Prove that $(a_1+1)(a_2+1) \cdots (a_n+1) +\frac{n}{4}(a^2_1+a^2_2+\cdots+a^2_n)\geq 1.$
Mongolia MO 2019 Grade 10 P3 wrote:

Let $a$ , $b$ , $c$ be real numbers such that $0\leq a\leq b\leq c$ . Prove the inequality
$(a+c)^2 (b+1)^2\geq (a+b+c+1)(ab+bc+ca+abc)$
Kazakhstan MO 2019 Grade 10 P1 wrote:

Let $a$ , $b$ , $c>0$ , such that $\tfrac{1}{a}+\tfrac{1}{b}+\tfrac{1}{c}=1$ . Prove the inequality
$\frac{b+c}{a+bc}+\frac{c+a}{b+ca}+\frac{a+b}{c+ab}\geq \frac{12}{a+b+c-1}$
RMO 2019 P3 wrote:

Let $a,b,c$ be positive real numbers such that $a+b+c=1$ . Prove that
$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\leq\frac{1}{5abc}$
Romania NMO 2019 Grade 9 wrote:

If $a,b,c\in(0,\infty)$ such that $a+b+c=3$ , then
$\frac{a}{3a+bc+12}+\frac{b}{3b+ca+12}+\frac{c}{3c+ab+12}\le \frac{3}{16}$
Kosovo MO 2019 Grade 10 P2 wrote:

Show that for any positive real numbers $a,b,c$ the following inequality is true:
$4(a^3+b^3+c^3+3)\geq 3(a+1)(b+1)(c+1)$ When does equality hold?
Indian Postals 2018 A P3 wrote:

Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=3$ . Prove that $\frac{1}{a}+\frac {3}{b}+\frac {5}{c}\ge 4a^2+3b^2+2c^2$ what does the equality hold?
ELMO SL 2019 A1 wrote:

Let $a$ , $b$ , $c$ be positive reals such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ . Show that $a^abc+b^bca+c^cab\ge 27bc+27ca+27ab.$

Proposed by Milan Haiman
Turkey TST 2019 P9 wrote:

Let $x, y, z$ be real numbers such that $y\geq 2z \geq 4x$ and $2(x^3+y^3+z^3)+15(xy^2+yz^2+zx^2)\geq 16(x^2y+y^2z+z^2x)+2xyz.$ Prove that: $4x+y\geq 4z$
Kyiv Math Festival 2019 wrote:

Let $a,b,c\ge0$ and $a+b+c\ge3.$ Prove that $a^4+b^3+c^2\ge a^3+b^2+c$
Serbia MO 2019 wrote:

Let $a,b,c\geq 0.$ Prove that $\sqrt{\frac{a+2b+3c}{3a+2b+c}}+\sqrt{\frac{b+2c+3a}{3b+2c+a}}+\sqrt{\frac{c+2a+3b}{3c+2a+b}}\geq 3$
HS Mathematics #2 Q610 wrote:

Let $x,y,z$ be positive real numbers. Prove that
$\frac{x^2(y+z)}{\sqrt{(z+x)(x+y)}}+\frac{y^2(z+x)}{\sqrt{(x+y)(y+z)}}+\frac{z^2(x+y)}{\sqrt{(y+z)(z+x)}} \geq xy+yz+zx.$
MEMO 2019 T1 wrote:

Determine the smallest and the greatest possible values of the expression
$\left( \frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right)\left( \frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)$ provided $a,b$ and $c$ are non-negative real numbers satisfying $ab+bc+ca=1$ .

Proposed by Walther Janous, Austria
Thailand MO 2019 P5 wrote:

Let $a,b,c$ be positive reals such that $abc=1$ . Prove the inequality
$\frac{4a-1}{(2b+1)^2} + \frac{4b-1}{(2c+1)^2} + \frac{4c-1}{(2a+1)^2}\geqslant 1.$
Romania NMO 2019 Grade 10 wrote:

If $a,b,c>0$ then
$\frac{1}{abc}+1\ge3\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{a+b+c}\right)$
Kvant 2019 #3 M2550 wrote:

Let $a,b,c>0$ be real numbers. Prove that
$\frac{a+b}{\sqrt{b+c}}+\frac{b+c}{\sqrt{c+a}}+\frac{c+a}{\sqrt{a+b}}\geq \sqrt{2a}+ \sqrt{2b}+ \sqrt{2c}$ Б. Кайрат (Казахстан), А. Храбров
China Jiangsu Summer Camp 2019 wrote:

Let $a$ , $b$ and $c$ be positive numbers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=a+b+c.$ Prove that
$\frac{1}{(2a+b+c)^2}+\frac{1}{(2b+c+a)^2}+\frac{1}{(2c+a+b)^2}\leq \frac{3}{16}.$
Slovenia TST 2019 P2 wrote:

Prove, that for any positive real numbers $a, b, c$ who satisfy $a^2+b^2+c^2=1$ the following inequality holds.
$\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c} \geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}$
CGMO Sichuan ST 2019 wrote:

Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=4.$ Prove that
$2\left(\frac{1}{ab}+\frac{1}{cd}\right)\geq a^2+b^2+c^2+d^2.$
CMO Summer Camp 2019 wrote:

Let $x,y,z$ be positive real number such that $xyz(x+y+z)=4.$ Find the minimum value of $(x+y)^2+2(y+z)^2+3(z+x)^2.$
Taiwan TST 2019 wrote:

Let $a,b,c,d$ be four non-negative reals such that $a+b+c+d = 4$ . Prove that $a\sqrt{3a+b+c}+b\sqrt{3b+c+d}+c\sqrt{3c+d+a}+d\sqrt{3d+a+b} \ge 4\sqrt{5}$
CHKMO 2018 P1 wrote:

Given that $a,b$ , and $c$ are positive real numbers such that $ab + bc + ca \geq 1$ , prove that
$\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \geq \frac{\sqrt{3}}{abc} .$
Greece 2018 wrote:

If $x, y, z$ are positive real numbers such that $x + y + z = 9xyz.$ Prove that:
$\frac {x} {\sqrt {x^2+2yz+2}}+\frac {y} {\sqrt {y^2+2zx+2}}+\frac {z} {\sqrt {z^2+2xy+2}}\ge 1.$
Consider when equality applies.
PAMO SL 2018 A6 wrote:

Let $a, b, c$ be positive real numbers such that $a^3 + b^3 + c^3 = 5abc$ .

Show that
$\left( \frac{a + b}{c} \right) \left( \frac{b + c}{a} \right) \left( \frac{c + a}{b} \right) \geq 9.$
Mongolia 2018 wrote:

Prove that for $a, b, c > 0$ such that $abc = 1$ following inequality holds $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{9}{2(a + b + c)} \geq \frac{9}{2}$
Romania NMO 2018 Grade 8 wrote:

Let $a$ , $b$ and $c$ be non-negative numbers such that $ab+ac+bc=3$ . Prove that:
$\frac{a}{a^2+7}+\frac{b}{b^2+7}+\frac{c}{c^2+7}\leq\frac{3}{8}.$
MEMO 2018 T1 wrote:

Let $a,b$ and $c$ be positive real numbers satisfying $abc=1.$ Prove that $\frac{a^2-b^2}{a+bc}+\frac{b^2-c^2}{b+ca}+\frac{c^2-a^2}{c+ab}\leq a+b+c-3.$
New-Zealand MO 2018 wrote:

Let $a,b,c$ be positive reals such that $\frac{1}{a+2019}+\frac{1}{b+2019}+\frac{1}{c+2019}=\frac{1}{2019}.$ Prove that $abc\geq{4038}^3.$
Estonia MO 2018 wrote:

$(1)$ Prove that for all real numbers $x, y, z$
$5 (x^2 + y^2 + z^2) \geq4 (xy + yz + zx).$ $(2)$ Prove that for any positive real number $x$
$(x + 1) (x + 2) (x + 5) \geq 36x.$
Balkan MO ShortList 2018 A4 wrote:

Let $a, b, c$ be positive real numbers such that $abc = 1.$ Prove that:
$2 (a^ 2 + b^ 2 + c^ 2) \left (\frac 1 {a^ 2} + \frac 1{b^ 2}+ \frac 1{c^2}\right)\geq 3(a+ b + c + ab + bc + ca).$
Balkan MO ShortList 2018 A1 wrote:

Let $a, b, c$ be positive real numbers such that $abc = \frac {2} {3}.$ Prove that:

$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$
St. Petersburg MO 2018 (Variant) wrote:

Let $a, b, c$ be positive numbers such that $a+b+c= 1.$ Show that
$\dfrac {a^ 2+b^ 2+c } {(a+b) ^3} +\dfrac {b^ 2+c^ 2+a} {(b +c) ^3} +\dfrac {c^ 2+a^ 2 +b } {( c+a) ^3}> 3$
Romania NMO 2018 Grade 9 P2 wrote:

Let $a,b,c \geq 0$ and $a+b+c=3.$ Prove that $\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \geq \frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{1+a}$
Moldova MO 2018 wrote:

For real numbers $a,b,c \in [0,1]$ prove that:
$\sum_{cyc}\frac{3a^2-2a}{1+b^3+c^3}\le1$
Beijing Girls Tournament 2018 wrote:

Let $a,b,c$ are positive real numbers . Prove that $a^3+b^3+c^3+2(ab^2+bc^2+ca^2)\geq3(a^2b+b^2c+c^2a).$
Turkey JBMO TST 2018 P8 wrote:

$x, y, z$ are positive real numbers and
$\sqrt {x}, \sqrt {y}, \sqrt {z}$ are lengths of a triangle such that :
$\frac {x} {y}+\frac {y} {z}+\frac {z} {x} =5$ .

Prove that: $\frac {x(y^2-2z^2)} {z}+\frac {y(z^2-2x^2)} {x} +\frac {z(x^2-2y^2)} {y} \geq 0$
HSGS TST 2018-19 wrote:

$x,y,z$ is non-negative such that: $\frac{x}{{x + 1}} + \frac{y}{{y + 1}} + \frac{z}{{z + 1}} = 1$ . Find minimum and maximum of: $P = xy + yz + zx + x\sqrt {yz} + y\sqrt {zx} + z\sqrt {xy}$
Croatia MO 2018 wrote:

Let $a, b$ and $c$ be positive real numbers such that $a+b+c=2.$ Prove that $\frac{(a-1)^2}{b}+\frac{(b-1)^2}{c}+\frac{(c-1)^2}{a}\geq \frac{1}{4}\left(\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\right).$
Vietnam 2018 wrote:

If $a, b, c$ are positive real numbers such that $\sqrt{a+2b} =2+\sqrt{\dfrac{b}{3}}$ . Prove that
$\frac{a}{{\sqrt {a + 2b} }} + \frac{b}{{\sqrt {b + 2a} }} \ge 2$
Dragon Boat Festival 2018 wrote:

$(3)$ Let $a,b,c;x,y,z\in[-1,1]$ and $1+2abc\geq a^2+b^2+c^2 , 1+2xyz\geq x^2+y^2+z^2.$ Prove that $1+2abcxyz\geq a^2x^2+b^2y^2+c^2z^2.$
Dragon Boat Festival 2018 wrote:

$(2)$ Let $a,b,c,d$ be positive real numbers .Prove that $\sqrt{\frac{a}{a+b+c}}+\sqrt{\frac{b}{b+c+d}}+\sqrt{\frac{c}{c+d+a}}+\sqrt{\frac{d}{d+a+b}}\leq\frac{4}{\sqrt{3}}.$
Dragon Boat Festival 2018 wrote:

$(1)$ Let $a_1,a_2,\cdots ,a_n$ be positive real numbers .Prove that $\sqrt{\frac{a_1}{a_1+a_2}}+\sqrt{\frac{a_2}{a_2+a_3}}+\cdots +\sqrt{\frac{a_n}{a_n+a_1}}\leq n-1.$
Mathematics Teaching #10 Q1041 wrote:

Let $a,b,c>0.$ Prove that $\left(1+\frac{a}{b} \right) \left(1+\frac{b}{c} \right) \left(1+\frac{c}{a} \right)+1\leq\frac{3(a^3+b^3+c^3)}{abc}.$
Delhi INMOTC 2018 wrote:

For all $a,b,c \in \mathbb{R}^+$ , Prove
$(a+b+c)^5 \geq 81abc(a^2+b^2+c^2)$
Macedonia JBMO TST 2018 wrote:

Let $x$ , $y$ , and $z$ be positive real numbers such that $x + y + z = 1$ . Prove that

$\frac{(x+y)^3}{z} + \frac{(y+z)^3}{x} + \frac{(z+x)^3}{y} + 9xyz \ge 9(xy + yz + zx)$ .

When does equality hold?
Moldova JBMO TST 2018 wrote:

Let $a,b,c \in\mathbb{R^*_+}$ .Prove the inequality $\frac{a^2+4}{b+c}+\frac{b^2+9}{c+a}+\frac{c^2+16}{a+b}\ge9$ .
Bulgaria JBMO TST 2018 P2 wrote:

For all positive reals $a$ and $b$ , show that
$\frac{a^2+b^2}{2a^5b^5} + \frac{81a^2b^2}{4} + 9ab > 18.$
Greece JBMO TST 2018 P1 wrote:

Let $a,b,c,d$ be non-negative real numbers such that $a^2+b^2+c^2+d^2=4$ . Prove that there exist two numbers with sum bigger or equal to $2$ .
FBH JBMO TST 2018 wrote:

1. For $a,b,c$ positive real numbers such that $ab+bc+ca=3$ prove:
$\frac{1}{1+a^2(b+c)}+\frac{1}{1+a^2(b+c)}+\frac{1}{1+a^2(b+c)} \leq \frac{a+b+c}{3abc}$
Moldova JBMO TST 2018 P6 wrote:

Let $a,b,c$ be positive real numbers such that $a+b+c=3$ . Show that
$\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\geq \frac{3}{2}.$
Slovenia TST 2018 P3 wrote:

Let $a$ , $b$ and $c$ be positive real numbers satisfying $abc=1$ . Prove that the following inequality holds:
$\frac{a+b+c}{3}\geq\frac{a}{a^2b+2}+\frac{b}{b^2c+2}+\frac{c}{c^2a+2}.$

This post has been edited 6 times. Last edited by AlastorMoody, Dec 3, 2019, 5:53 AM

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Alastor #proyaargodyaar
by GeoMetrix, Apr 24, 2020, 5:41 AM
@below why do you look so jobless
by AlastorMoody, Apr 16, 2020, 7:45 AM
e
eeeeeeeeeeeeeee
by gatorpokemon, Apr 15, 2020, 12:59 AM
Thanks a lot guyss and @below congrats for INMO merit
by AlastorMoody, Mar 19, 2020, 3:25 PM
Ahh! Now u belong to one of those kinds of ppl congrats!
by RAMUGAUSS, Mar 5, 2020, 7:28 PM
Congo for IMOTC!
by PhysicsMonster_01, Mar 3, 2020, 1:55 PM
Congrats alastor for IMOTC.
by GeoMetrix, Mar 3, 2020, 11:49 AM
im going to say something
by bingo2019, Feb 19, 2020, 1:43 AM
Probably because of JEE @below.
by gamerrk1004, Jan 29, 2020, 3:56 PM
Why are you living Olympiads Now ?
You have 2 more years in your high School ...
by a_simple_guy, Jan 20, 2020, 6:05 AM
contrib!!!
by Kagebaka, Jan 18, 2020, 9:48 PM
Hello
by A-student, Dec 19, 2019, 1:21 PM
hii.............................
by GeoMetrix, Nov 26, 2019, 12:50 PM
First shout man
by GAUSSIANGAUSS, Nov 16, 2019, 3:36 PM

14 shouts

Well, I am Dumb!!!!

I wanna help

by AlastorMoody, Dec 3, 2019, 5:23 AM

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