\( \newcommand{\N}{\mathbb{N}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\P}{\mathcal P} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathbb{F}} \newcommand{\E}{\mathcal E} \newcommand{\brac}[1]{\left(#1\right)} \newcommand{\abs}[1]{\left|#1\right|} \newcommand{\matrixx}[1]{\begin{bmatrix}#1\end {bmatrix}} \newcommand{\vmatrixx}[1]{\begin{vmatrix} #1\end{vmatrix}} \newcommand{\lims}{\mathop{\overline{\lim}}} \newcommand{\limi}{\mathop{\underline{\lim}}} \newcommand{\limn}{\lim_{n\to\infty}} \newcommand{\limsn}{\lims_{n\to\infty}} \newcommand{\limin}{\limi_{n\to\infty}} \newcommand{\nul}{\mathop{\mathrm{Nul}}} \newcommand{\col}{\mathop{\mathrm{Col}}} \newcommand{\rank}{\mathop{\mathrm{Rank}}} \newcommand{\dis}{\displaystyle} \newcommand{\spann}{\mathop{\mathrm{span}}} \newcommand{\range}{\mathop{\mathrm{range}}} \newcommand{\inner}[1]{\langle #1 \rangle} \newcommand{\innerr}[1]{\left\langle #1 \right \rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\toto}{\rightrightarrows} \newcommand{\upto}{\nearrow} \newcommand{\downto}{\searrow} \newcommand{\qed}{\quad \blacksquare} \newcommand{\tr}{\mathop{\mathrm{tr}}} \newcommand{\bm}{\boldsymbol} \newcommand{\cupp}{\bigcup} \newcommand{\capp}{\bigcap} \newcommand{\sqcupp}{\bigsqcup} \newcommand{\re}{\mathop{\mathrm{Re}}} \newcommand{\im}{\mathop{\mathrm{Im}}} \newcommand{\comma}{\text{,}} \newcommand{\foot}{\text{。}} \)

Tuesday, February 23, 2010

內地的高考

不等式是內地高考一個重要的課題。懷着好奇的心去做做看,不懂的還蠻多的...。以下都是些解了的題目,在此不提供解答了。

Problem 1. Prove that for any $  a,b,c\in\mathbb{R}$, $  \displaystyle \sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\ge \frac{3\sqrt{2}}{2}.$

Problem 2 (Mircea Lascu) Let $  a,b,c$ be positive real numbers such that $  abc=1$. Prove that $  \displaystyle \frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{2}}\ge \sqrt{a}+\sqrt{b}+\sqrt{c}+3.$

Problem 3. Let $  a,b,c,x,y,z$ be positive real numbers such that $  x+y+z=1$. Prove that $  \displaystyle ax+by+cz+2\sqrt{(xy+yz+zx)(ab+bc+ca)}\leq a+b+c.$


No comments:

Post a Comment