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

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Found problems: 14

Let $ a,b,c\in\mathbb{R}_+^*, $ and $ f:[0,a]\longrightarrow [0,b] $ bijective and non-decreasing. Prove that: $$ \frac{1}{b}\int_0^a f^2 (x)dx +\frac{1}{a}\int_0^b \left( f^{-1} (x)\right)^2dx\le ab. $$
Show that a continuous function $f : \mathbb{R} \to \mathbb{R}$ is increasing if and only if \[(c - b)\int_a^b f(x)\, \text{d}x \le (b - a) \int_b^c f(x) \, \text{d}x,\] for any real numbers $a < b < c$.
Let $\mathcal{C}$ be the set of all twice differentiable functions $f:[0,1] \to \mathbb{R}$ with at least two (not necessarily distinct) zeros and $|f''(x)| \le 1,$ for all $x \in [0,1].$ Find the greatest value of the integral $$\int\limits_0^1 |f(x)| \mathrm{d}x$$ when $f$ runs through the set $\mathcal{C},$ as well as the functions that achieve this maximum. [i]Note: A differentiable function $f$ has two zeros in the same point $a$ if $f(a)=f'(a)=0.$[/i]
Compute $$ \lim_{A\to+\infty}\frac1A\int_1^A A^{\frac1x}\, dx . $$ Proposed by Jan Šustek, University of Ostrava
Let $f: [0,\infty)\to \mathbb{R}$ a non-decreasing function. Then show this inequality holds for all $x,y,z$ such that $0\le x<y<z$. \begin{align*} & (z-x)\int_{y}^{z}f(u)\,\mathrm{du}\ge (z-y)\int_{x}^{z}f(u)\,\mathrm{du} \end{align*}
Let $f:\mathbb{R}\to\mathbb{R}^+$ be a continuous and periodic function. Prove that for all $\alpha\in\mathbb{R}$ the following inequality holds: $\int_0^T\frac{f(x)}{f(x+\alpha)}dx\ge T$, where $T$ is the period of $f(x)$.
Let $f:[0,1]\rightarrow R$ be a continuous function and n be an integer number,n>0.Prove that $\int_0^1f(x)dx \le (n+1)*\int_0^1 x^n*f(x)dx $
Suppose $f:\mathbb{R}\to \mathbb{R}$ is a two times differentiable function satisfying $f(0)=1,f^{\prime}(0)=0$ and for all $x\in [0,\infty)$, it satisfies \[ f^{\prime \prime}(x)-5f^{\prime}(x)+6f(x)\ge 0 \] Prove that, for all $x\in [0,\infty)$, \[ f(x)\ge 3e^{2x}-2e^{3x} \]
Prove that : \[\pi (e-1)<\int_0^{\pi} e^{|\cos 4x|}dx<2(e^{\frac{\pi}{2}}-1)\]
$f$ is a function twice differentiable on $[0,1]$ and such that $f''$ is continuous. We suppose that : $f(1)-1=f(0)=f'(1)=f'(0)=0$. Prove that there exists $x_0$ on $[0,1]$ such that $|f''(x_0)| \geq 4$
A function $f:[0,\infty)\to[0,\infty)$ is integrable and $$\int_0^\infty f(x)^2 dx<\infty,\quad \int_0^\infty xf(x) dx <\infty$$ Prove the following inequality. $$\left(\int_0^\infty f(x) dx \right)^3 \leq 8\left(\int_0^\infty f(x)^2 dx \right) \left(\int_0^\infty xf(x) dx \right)$$
Prove that for each positive integer $n$ \[\frac{4n^2+1}{4n^2-1}\int_0^{\pi} (e^{x}-e^{-x})\cos 2nx\ dx>\frac{e^{\pi}-e^{-\pi}-2}{4}\ln \frac{(2n+1)^2}{(2n-1)(n+3)}.\]
Prove that for each positive integer $n$ \[\frac{4n^2+1}{4n^2-1}\int_0^{\pi} (e^{x}-e^{-x})\cos 2nx\ dx>\frac{e^{\pi}-e^{-\pi}-2}{4}\ln \frac{(2n+1)^2}{(2n-1)(n+3)}.\]