Found problems: 592
2021 Taiwan TST Round 1, 4
Let $n$ be a positive integer. For each $4n$-tuple of nonnegative real numbers $a_1,\ldots,a_{2n}$, $b_1,\ldots,b_{2n}$ that satisfy $\sum_{i=1}^{2n}a_i=\sum_{j=1}^{2n}b_j=n$, define the sets
\[A:=\left\{\sum_{j=1}^{2n}\frac{a_ib_j}{a_ib_j+1}:i\in\{1,\ldots,2n\} \textup{ s.t. }\sum_{j=1}^{2n}\frac{a_ib_j}{a_ib_j+1}\neq 0\right\},\]
\[B:=\left\{\sum_{i=1}^{2n}\frac{a_ib_j}{a_ib_j+1}:j\in\{1,\ldots,2n\} \textup{ s.t. }\sum_{i=1}^{2n}\frac{a_ib_j}{a_ib_j+1}\neq 0\right\}.\]
Let $m$ be the minimum element of $A\cup B$. Determine the maximum value of $m$ among those derived from all such $4n$-tuples $a_1,\ldots,a_{2n},b_1,\ldots,b_{2n}$.
[I]Proposed by usjl.[/i]
2023 China Northern MO, 2
Let $ a,b,c \in (0,1) $ and $ab+bc+ca=4abc .$ Prove that $$\sqrt{a+b+c}\geq \sqrt{1-a}+\sqrt{1-b}+\sqrt{1-c}$$
1998 IMO Shortlist, 5
In a contest, there are $m$ candidates and $n$ judges, where $n\geq 3$ is an odd integer. Each candidate is evaluated by each judge as either pass or fail. Suppose that each pair of judges agrees on at most $k$ candidates. Prove that \[{\frac{k}{m}} \geq {\frac{n-1}{2n}}. \]
2018 ITAMO, 3
Let $x_1,x_2, ... , x_n$ be positive integers,Asumme that in their decimal representations no $x_i$ "prolongs" $x_j$.For instance , $123$ prolongs $12$ , $459$ prolongs $4$ , but $124$ does not prolog $123$.
Prove that :
$\frac {1}{x_1}+\frac {1}{x_2}+...+\frac {1}{x_n} < 3$.
2019 JBMO Shortlist, A2
Let $a, b, c $ be positive real numbers such that $abc = \frac {2} {3}. $ Prove that:
$$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$$
2025 6th Memorial "Aleksandar Blazhevski-Cane", P4
Prove that for all real numbers $a, b, c > 1$ the inequality
\[a(b^2 + c) + b(c^2 + a) + c(a^2 + b) \ge a^2 + b^2 + c^2 + 3abc\]
holds. When does equality hold?
Proposed by [i]Ilija Jovcevski[/i]
2021 China Team Selection Test, 4
Proof that
$$ \sum_{m=1}^n5^{\omega (m)} \le \sum_{k=1}^n\lfloor \frac{n}{k} \rfloor \tau (k)^2 \le \sum_{m=1}^n5^{\Omega (m)} .$$
2016 JBMO Shortlist, 1
Let $a, b, c$ be positive real numbers such that $abc = 8$. Prove that
$\frac{ab + 4}{a + 2}+\frac{bc + 4}{b + 2}+\frac{ca + 4}{c + 2}\ge 6$.
1990 IMO Shortlist, 24
Let $ w, x, y, z$ are non-negative reals such that $ wx \plus{} xy \plus{} yz \plus{} zw \equal{} 1$.
Show that $ \frac {w^3}{x \plus{} y \plus{} z} \plus{} \frac {x^3}{w \plus{} y \plus{} z} \plus{} \frac {y^3}{w \plus{} x \plus{} z} \plus{} \frac {z^3}{w \plus{} x \plus{} y}\geq \frac {1}{3}$.
1971 IMO, 1
Let \[ E_n=(a_1-a_2)(a_1-a_3)\ldots(a_1-a_n)+(a_2-a_1)(a_2-a_3)\ldots(a_2-a_n)+\ldots+(a_n-a_1)(a_n-a_2)\ldots(a_n-a_{n-1}). \] Let $S_n$ be the proposition that $E_n\ge0$ for all real $a_i$. Prove that $S_n$ is true for $n=3$ and $5$, but for no other $n>2$.
2009 IMO Shortlist, 2
Let $a$, $b$, $c$ be positive real numbers such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = a+b+c$. Prove that:
\[\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}.\]
[i]Proposed by Juhan Aru, Estonia[/i]
2024 Mongolian Mathematical Olympiad, 1
Let $P(x)$ and $Q(x)$ be polynomials with nonnegative coefficients. We denote by $P'(x)$ the derivative of $P(x)$. Suppose that $P(0)=Q(0)=0$ and $Q(1) \leq 1 \leq P'(0)$.
$(1)$ Prove that $0 \leq Q(x) \leq x \leq P(x)$ for all $0 \leq x \leq 1$.
$(2)$ Prove that $P(Q(x)) \leq Q(P(x))$ for all $0 \leq x \leq 1$.
[i]Proposed by Otgonbayar Uuye.[/i]
2013 Baltic Way, 4
Prove that the following inequality holds for all positive real numbers $x,y,z$:
\[\dfrac{x^3}{y^2+z^2}+\dfrac{y^3}{z^2+x^2}+\dfrac{z^3}{x^2+y^2}\ge \dfrac{x+y+z}{2}.\]
2013 Bosnia And Herzegovina - Regional Olympiad, 1
If $a$, $b$ and $c$ are nonnegative real numbers such that $a^2+b^2+c^2=1$, prove that $$\frac{1}{2} \leq \frac{a}{1+a^4}+\frac{b}{1+b^4}+\frac{c}{1+c^4} \leq \frac{9\sqrt{3}}{10}$$
1968 IMO Shortlist, 12
If $a$ and $b$ are arbitrary positive real numbers and $m$ an integer, prove that
\[\Bigr( 1+\frac ab \Bigl)^m +\Bigr( 1+\frac ba \Bigl)^m \geq 2^{m+1}.\]
2024 Junior Balkan Team Selection Tests - Romania, P1
For positive real numbers $x,y,z$ with $xy+yz+zx=1$, prove that
$$\frac{2}{xyz}+9xyz \geq 7(x+y+z)$$
2018 Balkan MO Shortlist, A1
Let $a, b, c $ be positive real numbers such that $abc = \frac {2} {3}. $ Prove that:
$$\frac {ab}{a + b} + \frac {bc} {b + c} + \frac {ca} {c + a} \geqslant \frac {a+b+c} {a^3+b ^ 3 + c ^ 3}.$$
1974 IMO Shortlist, 4
The sum of the squares of five real numbers $a_1, a_2, a_3, a_4, a_5$ equals $1$. Prove that the least of the numbers $(a_i - a_j)^2$, where $i, j = 1, 2, 3, 4,5$ and $i \neq j$, does not exceed $\frac{1}{10}.$
2010 Brazil Team Selection Test, 3
Let $a$, $b$, $c$ be positive real numbers such that $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = a+b+c$. Prove that:
\[\frac{1}{(2a+b+c)^2}+\frac{1}{(a+2b+c)^2}+\frac{1}{(a+b+2c)^2}\leq \frac{3}{16}.\]
[i]Proposed by Juhan Aru, Estonia[/i]
2014 239 Open Mathematical Olympiad, 6
Given posetive real numbers $a_1,a_2,\dots,a_n$ such that $a_1^2+2a_2^3+\dots+na_n^{n+1} <1.$ Prove that $2a_1+3a_2^2+\dots+(n+1)a_{n}^n <3.$
2019 India Regional Mathematical Olympiad, 3
Find all triples of non-negative real numbers $(a,b,c)$ which satisfy the following set of equations
$$a^2+ab=c$$
$$b^2+bc=a$$
$$c^2+ca=b$$
2021 Thailand TST, 2
Suppose that $a,b,c,d$ are positive real numbers satisfying $(a+c)(b+d)=ac+bd$. Find the smallest possible value of
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}.$$
[i]Israel[/i]
2002 Rioplatense Mathematical Olympiad, Level 3, 2
Let $\lambda$ be a real number such that the inequality $0 <\sqrt {2002} - \frac {a} {b} <\frac {\lambda} {ab}$ holds for an infinite number of pairs $ (a, b)$ of positive integers. Prove that $\lambda \geq 5 $.
2006 Federal Math Competition of S&M, Problem 1
Suppose $a,b,c,A,B,C$ are real numbers with $a\ne0$ and $A\ne0$ such that for all $x$,
$$\left|ax^2+bx+c\right|\le\left|Ax^2+Bx+C\right|.$$Prove that
$$\left|b^2-4ac\right|\le\left|B^2-4AC\right|.$$
2012 India Regional Mathematical Olympiad, 8
Let $x, y, z$ be positive real numbers such that $2(xy + yz + zx) = xyz$.
Prove that $\frac{1}{(x-2)(y-2)(z-2)} + \frac{8}{(x+2)(y+2)(z+2)} \le \frac{1}{32}$