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IMO练习曲(附:不等式选摘:Inequalities proposed in Crux Mathematicorum)

(2014-07-16 15:54:51)
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                                             IMOT 2011IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)                                       1998伊朗数学奥林匹克决赛试题

IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)
IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)
IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)
IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

                                                          problem  night 1

 http://www.artofproblemsolving.com/community/c6h599319p3557298

IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

                       Netherlands IMO Team Selection Tests 2014
                  IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)                 
                                                                 Dutch IMO TST I Problem 3
                     Spain Spain Mathematical Olympiad 2014                   IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)
IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

                         hoangthailelqd --arqady不等式           

        IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

                      mathwizarddude不等式

               IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum) 
IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)
 
 

IMO练习曲部分试题解答1

这段时间忙于装修,很少有连贯的时间来做题。

今天原计划去木工厂看家具,临出门了反悔。从十点多做到现在,解决了宋庆老师帖子:IMO练习曲(附:不等式选摘:Inequalities proposed in Crux Mathematicorum)里面十多道不等式,感觉有点饿了。

学习宋老师数学圈子里面的优良传统,选了三题的解答贴出来。

捕获

http://wuling.name/archives/2619
                                                           张云华:证等价
IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)
 
                                                 张云华:证一个三元代数不等式

 

IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

                                                      IMO练习曲
 
                                                          648 宋庆 IMO练习曲

参考文献     

1:宋庆 IMO练习曲

http://blog.sina.com.cn/s/blog_4c1131020102uxks.html

2熊昌进 2014西班牙奥林匹克第2题的证明

 http://blog.sina.com.cn/s/blog_64e168d30102uy07.html

 

                  不等式选摘:Inequalities proposed in Crux Mathematicorum   reny

【1528*】Let g(x)=\frac{x^2}{(x^2+1)^2} . Then the equation of the tangent line of g(x) at x=\frac{1}2} is given by y=t(x)=\frac{4}{25}+\frac{48}{125}\left(x-\frac{1}{2}\right) . To verify this, note that the slope of the tangent line at x=\frac{1}{2} is g'\left(\frac{1}{2}\right)=\frac{48}{125} , where g'(x)=\frac{2x(1-x^2)}{(x^2+1)^3} . As g\left(\frac{1}{2}\right) = \frac{4}{25} , we get the desired equation.

Now, consider the function f(x)=g(x)-t(x)=\frac{x^2}{(x^2+1)^2}-\frac{48x-4}{125} for x>0 . Then f(x)=-\frac{(2x-1)^2(12x^3+11x^2+32x-4)}{125(x^2+1)^2} . Note that the equation 12x^3+11x^2+32x-4=0 has exactly one positive real root x_0 < \frac{1}{8} by Descartes' Rule of Signs. Thus, f(x)\leq 0 for x\geq x_0 .

It follows that if \min\{a,b,c,d\}\geq x_0 and a+b+c+d=2 , then
\sum_{cyc} \frac{a^2}{(a^2+1)^2} \leq \sum_{cyc} \frac{48a-4}{125}=\frac{96-16}{125}=\frac{16}{25}.
Now, suppose that at least one of a,b,c,d is less than x_0 . WLOG, let a < x_0<\frac{1}{8} be fixed. Set t=\frac{2-a}{3} , so t\in\left(\frac{5}{8},\frac{2}{3}\right) .
Since g''(x)=\frac{2(3x^4-8x^2+1)}{(x^2+1)^4} , the point of inflection of g(x) is (c, g(c)) , where c=\sqrt{\frac{4-\sqrt{13}}{3}} , which is approximately 0.36 . Also, the tangent line at x=t is given by y=g(t)+g'(t)(x-t) . Since t > c , the tangent line at x=t is above the curve. Furthermore, one can verify that the tangent line at x=\frac{1}{\sqrt{3}} passes through the origin. Since t > \frac{1}{\sqrt{3}} , the tangent line at x=t is still above g(x) when x=0 . Therefore, we have g(x)\leq g(t)+g'(t)(x-t) . It follows that
\sum_{cyc} \frac{a^2}{(a^2+1)^2} \leq g(a)+3g(t)+g'(t)(b+c+d-3t)=g(a)+3g(t).
It suffices to show that h(a)<\frac{16}{25} , where h(a)=g(a)+3g\left(\frac{2-a}{3}\right) . We have
\begin{aligned}\frac{16}{25}-g(a)-3g\left(\frac{2-a}{3}\right)=\frac{4(2a-1)^2(a^6-7a^5-7a^4+56a^3-7a^2+263a+1)}{25(a^2+1)^2(...
and because p(x)=x^6-7x^5-7x^4+56x^3-7x^2+263x+1 has a root between 3 and 4 , with p(0)=1 and p(3)=700 , we have p(a)>0 for 0<a<\frac{1}{8} .
Therefore, the inequality is true and equality holds iff a=b=c=d=\frac{1}{2} .
 一个三元分式的最小值  reny

IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)
                            同一种方法解两题  一个三元不等式  reny IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

 

IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

 

IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

 

 

 

IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)

 

 

IMO练习曲(附:不等式选摘:Inequalities <wbr>proposed <wbr>in <wbr>Crux <wbr>Mathematicorum)


1. 已知 a,b,c 是正数. 求证:\sqrt[2]{ \frac{a^2+b^2+c^2}{3}} \le \frac{1}{3} \times (\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c} )                                         

                                                                                                                              problem 1 night 4      

 

2. 已知 a,b,c 是实数. 求证: \sqrt[3]{a} + \sqrt[3]{b} +\sqrt[3]{c} = \sqrt[3]{a+b+c} 等价于

                a^3 + b^3 + c^3 = (a + b + c)^3                                                               problem 7 night 3

3.已知 x,y,z > 1  且 \frac{1}{x} + \frac{1}{y} +\frac{1}{z} =2 求证: 

               \sqrt[2]{x+y+z} \ge \sqrt[2]{x-1}+\sqrt[2]{y-1}+\sqrt[2]{z-1}                                  problem 4 night 10

                                                                           1998伊朗数学奥林匹克决赛试题

4. 已知x,y 是满足 x^y + y = y^x + x 的正数 求证{ x+y \le 1 + xy

                                                                                                                              problem 7 night 8

5.已知x,y,z 是满足x*y*z = 1的正数 ,求证 :
               x^2 + y^2 + z^2 + xy + yz + zx \ge 2 \times (\sqrt[2]{x} + \sqrt[2]{y} + v\sqrt[2]{z})  
                                                                                                                              problem 1 night 5

6.已知a, b, c 两两不同, 均不为零,且a + b + c= 0. 求证: 

             ( \frac{b-c}{a} + \frac{c-a}{b} +\frac{a-b}{c})(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}) = 9                  problem 2 night 5

7.已知 a,b,c 为三角形三边长. 求证 :
a.  \sqrt[2]{a} , \sqrt[2]{b}, \sqrt[2]{c}. 为一三角形三边 长
b. \sqrt[2]{ab} + \sqrt[2]{bc} + \sqrt[2]{ca} \le a + b + c \le 2 * \sqrt[2]{ab} + 2* \sqrt[2]{bc} + 2* \sqrt[2]{ca}            problem 1 night 3

8.已知 a,b,c 为三角形三边长,且 a + b + c = 3 . 求a^2+b^2+c^2+ \frac{4abc}{3}  的最小值

                                                                                                                               problem 7 night  

9.已知 m n是非负整数,且  m > 1  , 2^{2m+1} \ge n^2 .求证2^{2m+1} \ge n^2 + 7   

                                                

                                                                                                                                problem 2 night 1

10.已知 p_1, p_2, \cdots , p_n(n \ge 2) 1, 2, 3,\cdots, n 的任意一个排列.求证:

    \frac{1}{p_1+p_2} + \frac{1}{p_2+p_3} +...+ \frac{1}{p_{n-1}+p_n} > \frac{n-1}{n+2}                                           problem 6 night 5

11.求证\sqrt{4n^2+n} 的小数部分不大于 \frac{1}{4}                                                                 problem 6 night 3
12.求证:\binom{2n}{n}< 4\sqrt{\frac{n}{3}} for n = 1, 2, 3, \cdots.                                                        problem 3 night 11
13.已知 a, b, c ∈ (1,2) 求证:

\frac{b\sqrt[2]{a}}{4b\sqrt[2]{c} - c\sqrt[2]{a}} + \frac{c\sqrt[2]{b}}{4c\sqrt[2]{a} - a\sqrt[2]{b}} + \frac{a\sqrt[2]{c}}{4...                                             problem  night 1
   已知ab  , c  是有理数,且 \frac{1}{a+bc}+\frac{1}{b+ac}=\frac{1}{a+b}. 证明\sqrt{(c-3)(c+1)} 是有理数.                        
                                                                                                                                     Dutch IMO TST I Problem 3

                                               Spain Spain Mathematical Olympiad 2014
 
                  1.   已知 rq  n 是满足\displaystyle\frac1{r+qn}+\frac1{q+rn}=\frac1{r+q} 的有理数, 证明\displaystyle\sqrt{\frac{n-3}{n+1}} 是一个有理数.                                                                                                                                            OME 2014 2
                   2.  已知 M 是一个形如a^2+13b^2的整数集合, 其中 a  , b 是不同的整数.

i) 证明 M 中任何两个元素的积仍是  M 的元素.

ii) Determine, reasonably, if there exist infinite pairs of integers (x,y) so thatx+y\not\in M but x^{13}+y^{13}\in M



                                                            hoangthailelqd --arqady不等式
             已知a,b,c\in (0;1) ,求证:a+b+c-ab-ac-bc<1.

                                                             mathwizarddude不等式

              已知 0\le x_1,x_2,\ldots,x_{10}\le\frac{\pi}{2} ,且\sin^2 x_1 +\sin^2 x_2+\cdots+\sin^2x_{10}=1 .求证
                                                            \frac{\cos x_1 +\cos x_2 +\cdots+\cos x_{10}}{\sin x_1+\sin x_2 +\cdots+\sin x_{10}} \ge 3  .
 
 

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