Found problems: 592
Let $a,b,c\geq 0$ and $a+b+c=1.$ Prove that$$\frac{a}{2a+1}+\frac{b}{3b+1}+\frac{c}{6c+1}\leq \frac{1}{2}.$$
[size=50](Marian Dinca)[/size]
Given real numbers $x_1,x_2,y_1,y_2,z_1,z_2$ satisfying $x_1>0,x_2>0,x_1y_1>z_1^2$, and $x_2y_2>z_2^2$, prove that: \[ {8\over(x_1+x_2)(y_1+y_2)-(z_1+z_2)^2}\le{1\over x_1y_1-z_1^2}+{1\over x_2y_2-z_2^2}. \] Give necessary and sufficient conditions for equality.
For all integers $ {a}_{0},{a}_{1}, \cdot\cdot\cdot {a}_{100}$, find the maximum of ${a}_{5}-2{a}_{40}+3{a}_{60}-4{a}_{95} $
$\bigstar$ 1) ${a}_{0}={a}_{100}=0$
2) for all $i=0,1,\cdot \cdot \cdot 99, $ $|{a}_{i+1}-{a}_{i}|\le1$
3) $ {a}_{10}={a}_{90} $
A sequence $x_1, x_2, \ldots$ is defined by $x_1 = 1$ and $x_{2k}=-x_k, x_{2k-1} = (-1)^{k+1}x_k$ for all $k \geq 1.$ Prove that $\forall n \geq 1$ $x_1 + x_2 + \ldots + x_n \geq 0.$
[i]Proposed by Gerhard Wöginger, Austria[/i]
Show that if $a, b, c$ are the lengths of the sides of a triangle and if $2S = a + b + c$, then
\[\frac{a^n}{b+c} + \frac{b^n}{c+a} +\frac{c^n}{a+b} \geq \left(\dfrac 23 \right)^{n-2}S^{n-1} \quad \forall n \in \mathbb N \]
[i]Proposed by Greece.[/i]
Prove that
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \ge \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+2(a+b+c)$$
for the all $a,b,c$ positive real numbers satisfying $a^2+b^2+c^2+2abc \le 1$.
For real numbers $0 \leq a_1 \leq a_2 \leq ... \leq a_n$ and $0 \leq b_1 \leq b_2 \leq ... \leq b_n$ prove that \[ \left( \frac{a_1}{1 \cdot 2}+\frac{a_2}{2 \cdot 3}+...+\frac{a_n}{n(n+1)} \right) \times \left( \frac{b_1}{1 \cdot 2}+\frac{b_2}{2 \cdot 3}+...+\frac{b_n}{n(n+1)} \right) \leq \frac{a_1b_1}{1 \cdot 2}+\frac{a_2b_2}{2 \cdot 3}+...+\frac{a_nb_n}{n(n+1)}.\]
[i]Proposed by A. Antropov[/i]
Let $a$, $b$, $c$, $d$ be real numbers satisfying
\begin{align*}
(a + c)(b + d) = \sqrt{2}(ac - 2bd - 1).
\end{align*}
Show that
\begin{align*}
(ab - 1)^2 + (bc - 1)^2 + (cd - 1)^2 + (da - 1)^2 + (ac - 1)^2 + (2bd + 1)^2 \ge 4.
\end{align*}
$\boxed{\text{A5}}$ Let $x,y,z$ be positive reals. Prove that $(x^2+y+1)(x^2+z+1)(y^2+x+1)(y^2+z+1)(z^2+x+1)(z^2+y+1)\geq (x+y+z)^6$
Prove that, for all pairs of nonnegative integers, $j,n$, $$\sum_{K=0}^{n}k^j\binom n k \ge 2^{n-j} n^j$$
Let $f$ be an injective function from ${1,2,3,\ldots}$ in itself. Prove that for any $n$ we have: $\sum_{k=1}^{n} f(k)k^{-2} \geq \sum_{k=1}^{n} k^{-1}.$
Prove that for all non-negative real numbers $x,y,z$, not all equal to $0$, the following inequality holds
$\displaystyle \dfrac{2x^2-x+y+z}{x+y^2+z^2}+\dfrac{2y^2+x-y+z}{x^2+y+z^2}+\dfrac{2z^2+x+y-z}{x^2+y^2+z}\geq 3.$
Determine all the triples $(x,y,z)$ for which the equality holds.
[i]Milan Mitreski, Serbia[/i]
Let $a,b,c$ be positive real numbers such that $a+b+c=1$. Prove that
$$\frac{a}{a^2+b^3+c^3}+\frac{b}{b^2+a^3+c^3}+\frac{c}{c^2+a^3+b^3}\leq\frac{1}{5abc}$$
The positive integers $x_1, \cdots , x_n$, $n \geq 3$, satisfy $x_1 < x_2 <\cdots< x_n < 2x_1$. Set $P = x_1x_2 \cdots x_n.$ Prove that if $p$ is a prime number, $k$ a positive integer, and $P$ is divisible by $pk$, then $\frac{P}{p^k} \geq n!.$
Prove inequality, with $a$ and $b$ nonnegative real numbers:
$\frac{a+b}{1+a+b}\leq \frac{a}{1+a} + \frac{b}{1+b} \leq \frac{2(a+b)}{2+a+b}$
Show that if $a, b, c$ are the lengths of the sides of a triangle and if $2S = a + b + c$, then
\[\frac{a^n}{b+c} + \frac{b^n}{c+a} +\frac{c^n}{a+b} \geq \left(\dfrac 23 \right)^{n-2}S^{n-1} \quad \forall n \in \mathbb N \]
[i]Proposed by Greece.[/i]
Let $ \lfloor x \rfloor$ denote the greatest integer less than or equal to $ x.$ Pick any $ x_1$ in $ [0, 1)$ and define the sequence $ x_1, x_2, x_3, \ldots$ by $ x_{n\plus{}1} \equal{} 0$ if $ x_n \equal{} 0$ and $ x_{n\plus{}1} \equal{} \frac{1}{x_n} \minus{} \left \lfloor \frac{1}{x_n} \right \rfloor$ otherwise. Prove that
\[ x_1 \plus{} x_2 \plus{} \ldots \plus{} x_n < \frac{F_1}{F_2} \plus{} \frac{F_2}{F_3} \plus{} \ldots \plus{} \frac{F_n}{F_{n\plus{}1}},\]
where $ F_1 \equal{} F_2 \equal{} 1$ and $ F_{n\plus{}2} \equal{} F_{n\plus{}1} \plus{} F_n$ for $ n \geq 1.$
Let $ a,b,c \in (0,1) $ and $ab+bc+ca=4abc .$ Prove that $$\sqrt{a+b+c}\geq \sqrt{1-a}+\sqrt{1-b}+\sqrt{1-c}$$
Find all real values of $K$ which satisfies the following.
Let there be a sequence of real numbers $\{a_n\}$ which satisfies the following for all positive integers $n$.
(i). $0 < a_n < n^K$.
(ii). $a_1 + a_2 + \cdots + a_n < \sqrt{n}$.
Then, there exists a positive integer $N$ such that for all integers $n>N$, $$a^{2018}_1 + a^{2018}_2 + \cdots +a^{2018}_n < \frac{n}{2018}$$
Let $a$, $b$, $c$, $k$ be positive reals such that $ab+bc+ca \leq 1$ and $0 < k \leq \frac{9}{2}$. Prove that:
\[\sqrt[3]{ \frac{k}{a} + (9-3k)b} + \sqrt[3]{\frac{k}{b} + (9-3k)c} + \sqrt[3]{\frac{k}{c} + (9-3k)a } \leq \frac{1}{abc}.\]
[i]Proposed by Zhang Yanzong and Song Qing[/i]
Given an integer $n\ge3$ consider positive integers $x_1,\ldots,x_n$ such that $x_1<x_2<\cdots<x_n<2x_1$. If $p$ is a prime and $r$ is a positive integer such that $p^r$ divides the product $x_1\cdots x_n$, prove that $$\frac{x_1\cdots x_n}{p^r}>n!.$$
Let $M(t)$ be measurable and locally bounded function, that is,
\[M(t) \le C_{a,b}, \quad \forall 0 \le a \le t \le b<\infty\]
with some constant $C_{a,b}$, from $[0,\infty)$ to $[0,\infty)$ such that
\[M(t) \le 1+\int_0^t M(t-s)(1+t)^{-1}s^{-1/2} ds, \quad \forall t \ge 0.\]
Show that
\[M(t) \le 10+2\sqrt{5}, \quad \forall t \ge 0.\]
Let $Z^+$ be positive integers set. $f:\mathbb{Z^+}\to\mathbb{Z^+}$ is a function and
we show $ f \circ f \circ ...\circ f $ with $f_l$ for all $l\in \mathbb{Z^+}$ where $f$ is repeated $l$ times. Find all $f:\mathbb{Z^+}\to\mathbb{Z^+}$ functions such that $$ (n-1)^{2020}< \prod _{l=1}^{2020} {f_l}(n)< n^{2020}+n^{2019} $$
for all $n\in \mathbb{Z^+}$
$a,b,c$ are nonnegative integers that satisfy $a^2+b^2+c^2=3$. Find the minimum and maximum value the sum
$$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}$$
may achieve and find all $a,b,c$ for which equality occurs.\\ \\
[i](Andrei Bâra)[/i]
Given a real number $C\leqslant 2$. Prove that for every positive real number $x,y$ with $xy=1$, the following inequality holds:
\[ \sqrt{\frac{x^2+y^2}{2}} + \frac{C}{x+y} \geqslant 1 + \frac{C}{2}.\]
[i]Proposed by Fajar Yuliawan, Indonesia[/i]