This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

Tags were heavily modified to better represent problems.

AND
OR
NO

Found problems: 10

$a)$ Prove that $2x^3-3x^2+1\geq 0,~(\forall)x\geq0.$ $b)$ Let $x,y,z\geq 0$ such that $\frac{2}{1+x^3}+\frac{2}{1+y^3}+\frac{2}{1+z^3}=3.$ Prove that $\frac{1-x}{1-x+x^2}+\frac{1-y}{1-y+y^2}+\frac{1-z}{1-z+z^2}\geq 0.$
Let $n > 3$ be a natural number, and let $x_1$, $x_2$, ..., $x_n$ be $n$ positive real numbers whose product is $1$. Prove the inequality \[ \frac {1}{1 + x_1 + x_1\cdot x_2} + \frac {1}{1 + x_2 + x_2\cdot x_3} + ... + \frac {1}{1 + x_n + x_n\cdot x_1} > 1. \]
Suppose $a,b,c\in \mathbb R^+$. Prove that :\[\left(\frac ab+\frac bc+\frac ca\right)^2\geq (a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\]
If $a, b, c, d$ are positive numbers, prove that $$\sum_{cyclic}\frac{a-\sqrt[3]{bcd}}{a+3(b+c+d)}\ge 0$$
Find the smallest constant $ k$ such that $ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}}\leq k\sqrt {x \plus{} y \plus{} z}$ for all positive $ x$, $ y$, $ z$.
Find the smallest constant $ k$ such that $ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}}\leq k\sqrt {x \plus{} y \plus{} z}$ for all positive $ x$, $ y$, $ z$.
Given real numbers $a,b,c,d,e>1$. Prove that \[ \frac{a^2}{c-1}+\frac{b^2}{d-1}+\frac{c^2}{e-1}+\frac{d^2}{a-1}+\frac{e^2}{b-1} \ge 20 \]
Let $a,b,c$ be the positive real numbers satisfying $a^2+b^2+c^2=3$. Prove that: $$\frac{a}{b(a+c)}+\frac{b}{c(b+a)}+\frac{c}{a(c+b)}\geq \frac{3}{2}.$$
Prove that for all positive real numbers $a,b,c$ the following inequality takes place \[ \frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)} \geq \frac{27}{2(a+b+c)^2} . \] [i]Laurentiu Panaitopol, Romania[/i]
Let $a$, $b$, $c \in (0,+\infty)$ . Then the following inequality is true:$$\sqrt{(a+b)(b+c)}+\sqrt{(b+c)(c+a)}+\sqrt{(c+a)(a+b)}+a+b+c\leq \left(ab+bc+ca\right)\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\right)$$