Found problems: 6
Let $D$ be the interior of the circle $C$ and let $A \in C$. Show that the function $f : D \to \mathbb R, f(M)=\frac{|MA|}{|MM'|}$ where $M' = AM \cap C$, is strictly convex; i.e., $f(P) <\frac{f(M_1)+f(M_2)}{2}, \forall M_1,M_2 \in D, M_1 \neq M_2$ where $P$ is the midpoint of the segment $M_1M_2.$
Let $I$ and $J$ be intervals. Let $\varphi,\psi:I\to\mathbb{R}$ be strictly increasing continuous functions and let $\Phi,\Psi:J\to\mathbb{R}$ be continuous functions. Suppose that $\varphi(x)+\psi(x)=x$ and $\Phi(u)+\Psi(u)=u$ holds for all $x\in I$ and $u\in J$. Show that if $f:I\to J$ is a continuous solution of the functional inequality
$$f\big(\varphi(x)+\psi(y)\big)\le \Phi\big(f(x)\big)+\Psi\big(f(y)\big)\qquad (x,y\in I),$$then $\Phi\circ f\circ \varphi^{-1}$ and $\Psi\circ f\circ \psi^{-1}$ are convex functions.
Let $f : (0,+\infty) \to \mathbb{R}$ a convex function and $\alpha, \beta, \gamma > 0$. Then $$\frac{1}{6\alpha}\int \limits_0^{6\alpha}f(x)dx\ +\ \frac{1}{6\beta}\int \limits_0^{6\beta}f(x)dx\ +\ \frac{1}{6\gamma}\int \limits_0^{6\gamma}f(x)dx$$ $$\geq \frac{1}{3\alpha +2\beta +\gamma}\int \limits_0^{3\alpha +2\beta +\gamma}f(x)dx\ +\ \frac{1}{\alpha +3\beta +2\gamma}\int \limits_0^{\alpha +3\beta +2\gamma}f(x)dx\ $$ $$+\ \frac{1}{2\alpha +\beta +3\gamma}\int \limits_0^{2\alpha +\beta +3\gamma}f(x)dx$$
Let $f:\mathbb{Q}\times\mathbb{Q}\to\mathbb{Q}$ be a function satisfying:
[list]
[*] For any $x_1,x_2,y_1,y_2 \in \mathbb Q$, $$f\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right) \leq \frac{f(x_1,y_1)+f(x_2,y_2)}{2}.$$
[*] $f(0,0) \leq 0$.
[*] For any $x,y \in \mathbb Q$ satisfying $x^2+y^2>100$, the inequality $f(x,y)>1$ holds.\
Prove that there is some positive rational number $b$ such that for all rationals $x,y$, $$f(x,y) \ge b\sqrt{x^2+y^2} - \frac{1}{b}.$$
For those real numbers $x_1 , x_2 , \ldots , x_{2011}$ where each of which satisfies $0 \le x_1 \le 1$ ($i = 1 , 2 , \ldots , 2011$), find the maximum of
\[ x_1^3+x_2^3+ \cdots + x_{2011}^3 - \left( x_1x_2x_3 + x_2x_3x_4 + \cdots + x_{2011}x_1x_2 \right) \]
Let $D$ be the interior of the circle $C$ and let $A \in C$. Show that the function $f : D \to \mathbb R, f(M)=\frac{|MA|}{|MM'|}$ where $M' = AM \cap C$, is strictly convex; i.e., $f(P) <\frac{f(M_1)+f(M_2)}{2}, \forall M_1,M_2 \in D, M_1 \neq M_2$ where $P$ is the midpoint of the segment $M_1M_2.$