标签：cal must keep 结果 phi $? triple can max
记得17岁那年 第一次和她接吻 快亲上的时候 她突然说等一下 我就纳闷了 她要干嘛？只见她小心翼翼地从兜里拿出三个糖 就上好佳那种圆的 草莓苹果和荔枝味的 她让我挑一个喜欢的 我指了一下那个荔枝的 然后问她干嘛? 她二话不说马上撕开糖纸 就把那颗糖给吃了 然后一把扯过我的脖子 我俩就接吻了 全程一股荔枝味 后来她跟我说 人生那么长 我没有自信能让你记住我 但是你既然喜欢吃荔枝味的糖 我只能让你记住 我和你接吻是荔枝味的 这样以后你吃荔枝味的东西都能想起我 我和你接吻的味道。 如今我们分手好多年了 每次吃荔枝味的东西都会想起她 家里固定有荔枝糖 想她了都会吃上一个 就好像在和她接吻。若还有机会 真想告诉她 人生那么长 我可能要记着你一辈子了。后来 我有过两个女友 也终没有结果 时间就这样沉淀下去 终于有一天 我再也无法抑制我心中的那份情感 我决定去找她 我们要在一起 后来经多方打听才知道 她毕业后找了份不错的工作 工作几年后 毅然辞职自己开了家糖果店 并经过努力打拼 慢慢积累了一些财富 而我终于有一天找到她 开口的第一句： 还记得那次荔枝糖的味道吗？ 她强忍着泪告诉我 荔枝糖的味道她一直没忘记 只是我们再也回不去 了。我没有转身离开，也没有奋不顾身的冲上去抱住她说出多年来心里一直只想对她说的那些话。就这样，我们傻傻地看着对方，彼此沉默了很久。夕阳的余晖透过窗户斜映在她的脸庞，一如当年那般楚楚动人，突然心里流过一股暖意，仿佛那些年曾一起走过的旧时光还在身旁。或许，这已经足够了。有些人，有些事，一旦错过了就是错过，不再擦肩，也不再回头。虽然，岁月带走了我心中最美好的曾经，但是我并不会很伤心，因为，这都是我瞎编的！”
意哥负责（3，6，7，8，9，10，19，34）
姜老师负责几何（1，4，11，23，24，26，27）
熊哥负责不等式
逆逆负责数论
剩下的有：（2，5，12，13，14，15，16，17，18，20，21，22，25，28，29，30，31，32，33）
60和66再检查！！！
设$f(x)$在$\mathbb{R}^m$上连续,对$x\neq 0$有$f(x)>0$,且对任意$x\in\mathbb{R}^m$和$c>0$有$f(cx)=cf(x)$.求证存在$a>0$和$b>0$使得\[a|x|\leq f(x)\leq b|x|.\]
证:显然$c\neq 1$,否则考察$f(x)=|x|^2$.首先令$x=0$,则我们有$|f(0)|=|f(c0)|=c|f(0)|$,则$f(0)=0$.
当$x\neq 0$时,我们有
$$\frac{f\left( c^nx \right)}{c^n\left| x \right|}=\frac{cf\left( c^{n-1}x \right)}{c^n\left| x \right|}=\frac{f\left( c^{n-1}x \right)}{c^{n-1}\left| x \right|}=\cdots =\frac{f\left( x \right)}{\left| x \right|}$$
及
$$\frac{f\left( x \right)}{\left| x \right|}=\frac{f\left( x/c^n \right)}{\left| x \right|/c^n},$$
当$0<c<1$时,取足够大的$n$使得$r_1\leq c^n\left| x \right|\leq r_2$,并记$$l_1=\sup_{\left| x \right|<r_2}\left| f\left( x \right) \right|,\quad l_2=\inf_{r_1\leq \left| x \right|\leq r_2}\left| f\left( x \right) \right|,$$我们有
$$\frac{l_2}{r_2}\leq \frac{f\left( x \right)}{\left| x \right|}=\frac{f\left( c^nx \right)}{c^n\left| x \right|}\leq \frac{l_1}{r_1},\quad x\neq 0$$
取$a=l_2/r_2,b=l_1/r_1$,便有$a|x|\leq f(x)\leq b|x|$对$x\neq0$成立,显然也对$x=0$成立.若$c>1$,考察
$$\frac{f\left( x \right)}{\left| x \right|}=\frac{f\left( x/c^n \right)}{\left| x \right|/c^n}$$类似地可以证得.
\section{Proving Inequalities Using Linear Functions}%12
\markboth{Articles}{Proving Inequalities Using Linear Functions}
\vspace{4.2cm}
In this article we present a method for proving a class of inequalities based
on the simple observation that if a linear real function attains values of the
same sign at the end points of an interval, all of its values are of the same sign
on the whole interval.
For this purpose, it is crucial to view an expression as
a linear function in certain group of the variables.
\begin{center}
\includegraphics{42.pdf}
\end{center}
\bc
Figure 3.12-1
\ec
{\bf Theorem 1.}
{\it If the function $f(x)=ax+b$ has $f(\alpha )\ge 0$ and $f(\beta )\ge 0$
then $f(x)\ge 0$, $\forall \ x\in [\alpha ,\beta ]$.
}
This property of linear functions is well illustrated in Figure 3.12-1, and it
has an easily understood geometric interpretation.
We will illustrate the idea with two problems.
{\bf Problem 1.}
{\it Let $x,y,z$ be nonnegative real numbers such that $x+y+z=3$.
Prove that
$$x^2+y^2+z^2+xyz\ge 4.$$
}
{\bf Solution.}
We write the desired inequality in the form
$$(y+z)^2-2yz+x^2+xyz\ge 4,$$
or
$$yz(x-2)+2x^2-6x+5\ge 0.$$
Let $yz = w$, and view the expression on the left-hand side, as a linear
function of $w$, that is
$$f(w)=(x-2)w+2x^2-6x+5.$$
Now we need to find all possible values of $w$.
By the AM-GM inequality,
$yz\le (y+z)^2/4$.
That is
$$w\le (3-x)^2/4,$$
where $w\ge 0$ by hypothesis.
By Theorem 1, it is sufficient to prove that
$f(0)\ge 0$ and $f(w_0)\ge 0$, where $w_0=(3-x)^2/4$.
It is not difficult to check that
$$f(0)=2x^2-6x+5=2\left(x-\ds\f{3}{2}\right)^2+\ds\f{1}{5}\ge 0,$$
$$f(w_0)=\ds\f{1}{4}(x-1)^2(x+2)\ge 0.$$
The proof is complete.
The equality holds if and only if all three numbers are equal to 1.
Actually, in order to determine the equality cases we find all sets of values
of the variables so that the values of the linear function at endpoints of interval
in question are all zeros.
For instance, in Problem 1, equation $f(0) = 0$ has no real solution.
While we can find out that $f(w_0) = 0$ has a root $x = 1$, which leads to
$yz\le 1$, $y+z=2$.
Plugging $z=2-y$ into the inequality $yz\le 1$
we obtain
$$-y^2+2y-1\le 0.$$
This yields $y = 1$, then we have $z = 1.$
The next problem has two equality cases determined by solving two equations
$f(0) = 0$ and $f(w_0) = 0$.
This is also the case when you solve Schur‘s
inequality in three variables (the last exercise).
{\bf Problem 2.}
{\it Prove that if $x,y,z$ are non-negative real numbers such that
$x+y+z=1$, then
$$4(x^3+y^3+z^3)+15xyz\ge 1.$$
Determine when equality holds.
}
{\bf Solution.}
Note that we have the following identity
$$a^3+b^3=(a+b)^3-3ab(a+b),$$
so that the given inequality is equivalent to
$$(y+z)^3-3yz(y+z)+x^3+\ds\f{15}{4}xyz\ge \ds\f{1}{4}.$$
Using the fact that the three numbers add up to 1, the above inequality reads
$$(1-x)^3+yz\left(\ds\f{27}{4}x-3\right)+x^3-\ds\f{1}{4}\ge 0.$$
Let $yz = w$ and consider the left-hand side of the inequality as a linear function
of $w$
$$f(w)=\left(\ds\f{27}{4}x-3\right)w+(1-x)^3+x^3-\ds\f{1}{4}.$$
By the AM-GM inequality,
$w\le (1-x)^2/4$, and we also have
$w\ge 0$ by hypothesis.
By Theorem 1, we will prove that $f(0)\ge 0$ and $f(w_0)\ge 0$.
Indeed, we have
$$f(0)=(1-x)^3+x^3-\ds\f{1}{4}=\ds\f{3}{4}(2x-1)^2,$$
$$16f(w_0)=16(1-x)^3+(1-x)^2(27x-12)+16x^3-4=3x(3x-1)^2.$$
The given inequality follows, and the proof is complete.
Equality occurs if
$(x,y,z)=\left(\ds\f{1}{3},\ds\f{1}{3},\ds\f{1}{3}\right)$,
or for any permutation of the triple
$(x,y,z)=\left(0,\ds\f{1}{2},\ds\f{1}{2}\right)$.
{\bf Exercise 1.}
(Mihai Piticari, Dan Popescu, Old \& New Inequalities).
Prove that
$$5(a^2+b^2+c^2)\le 6(a^3+b^3+c^3)+1,$$
where $a,b,c$ are positive real numbers such that $a + b + c = 1$.
{\bf Exercise 2.}
(Sefket Arslanagic, Crux Maths).
Prove the inequality
$$\ds\f{1}{1-xy}+\ds\f{1}{1-yz}+\ds\f{1}{1-zx}\le \ds\f{27}{8},$$
where $x,y,z$ are positive real numbers such that $x + y + z = 1$.
{\bf Exercise 3.}
(BMO 1979).
Let $x,y,z$ be positive real numbers such that $x +y + z = 1$.
Prove that
$$7(xy+yz+zx)\le 2+9xyz.$$
{\bf Exercise 4.}
(USAMO 1979).
Prove that if $x,y,z>0$ and $x+y+z=1$, then
$$x^3+y^3+z^3+6xyz\ge 1/4.$$
{\bf Exercise 5.}
(IMO 1984).
Prove that if $x,y,z>0$ and $x+y+z=1$, then
$$xy+yz+zx-2xyz\le 7/27.$$
{\bf Exercise 6.}
(I. Schur).
Prove that if $a,b,c\ge 0$, then
$$a^3+b^3+c^3+3abc\ge a^2(b+c)+b^2(c+a)+c^2(a+b).$$
{\bf Acknowledgement.}
The first author thanks Pham Huu Duc (Australia) for
useful suggestions in exposition.
\section*{Bibliography}
\bi
\item[{[1]}]
T. Andrescu, V. Cartoaje, G. Dospinescu, M. Lascu,
{\it Old and New Inequalities},
GIL Publishing House, 2004.
\item[{[2]}]
P.V. Thuan, D. Grinberg, H.N. Minh,
{\it Inequalities,}
Reasoning and Exploration, 2007.
\item[{[3]}]
S. Arslanagic,
{\it Problem 2876},
Crux Mathematicorum, Canadian Mathematical Society, 2002.
\ei
\bigskip
\hfill
{\Large Pham Van Thuan and Trieu Van Hung, Hanoi, Vietnam}
\section{A Note on the Breaking Point of a Simple Inequality}%14
\markboth{Articles}{A Note on the Breaking Point of a Simple Inequality}
\vspace{4.2cm}
Let us start with the well-known Mongolian inequality
$$\left(\ds\f{x+y+z}{3}\right)^3\ge \ds\f{(x+y)(y+z)(z+x)}{8}.$$
Another inequality which bears some similarity to it,
$$\left(\ds\f{x+y+z}{3}\right)\left(\ds\f{xy+yz+zx}{3}\right)
\le \ds\f{(x+y)(y+z)(z+x)}{8},$$
is also popular in problem solving communities.
We can see that the direction is reversed in the second inequality.
So it occurred to the authors to find the limit point where the inequality is still true.
This is the problem we will solve in this article.
{\bf Problem.}
{\it Find the minimum value for a such that the inequality
$$\left(\ds\f{x+y+z}{3}\right)^a \left(\ds\f{xy+yz+zx}{3}\right)^\f{3-a}{2}
\ge \ds\f{(x+y)(y+z)(z+x)}{8}
\eqno(1)$$
holds for all positive $x,y,z$.
}
{\bf Solution.}
Let us try to maximize the right-hand side keeping the left-hand
side fixed.
Let $m = x + y + z$, $n = xy + yz + zx$, $p = xyz$.
We rewrite the left-hand side as
$\left(\ds\f{m}{3}\right)^a \left(\ds\f{n}{3}\right)^\f{3-a}{2}$
and the right-hand side as
$\ds\f{mn-p}{8}$.
So keeping $m,n$
fixed and decreasing $p$ will keep the left-hand side fixed but will increase
the right-hand side.
Now how small can we make $p$?
We must ensure that the equation
$x^3-mx^2+nx-p=0$
has three positive real solutions, or that the
line $y = p$ intersects the graph of
$y=x^3-mn^2+nx$
in three points to the right of the $y$-axis.
Therefore the extremal case is either when the line intersects
the graph on the $y$-axis (so that moving further would produce a point of
intersection to the left of the $y$-axis, a negative solution) which means one
of the variables is zero, or when the line touches the graph (because moving
further would yield fewer points of intersections), which is equivalent to two
variables being equal.
Therefore it suffices to look at these cases one by one.
{\bf Case 1.}
Assume that $z = 0$.
Because the inequality is symmetric and homogeneous, we can assume $y = 1$.
The inequality becomes,
$$\left(\ds\f{x+1}{3}\right)^a \left(\ds\f{x}{3}\right)^\f{3-a}{2}
\ge \ds\f{(x+1)x}{8}.
\eqno(2)$$
Squaring, we get
$64(x+1)^{2a}x^{3-a}\ge 3^{3+a}(x+1)^2 x^2$,
or
$64(x+1)^{2(a-1)}\ge 3^{3+a}x^{a-1}$,
which is equivalent to
$$\left[\ds\f{(x+1)^2}{3x}\right]^{a-1}\ge \ds\f{81}{64}.$$
As $\ds\f{(x+1)^2}{3x}$
can take values as large as we want, we conclude that $a-1>0$.
Then the minimal value of
$\ds\f{(x+1)^2}{3x}$ is $\ds\f{4}{3}$
which occurs for $x = 1$, so we must compute $a$ for $x = 1$, which gives us
$$\left(\ds\f{4}{3}\right)^{a-1}\ge \ds\f{81}{64}$$
$$a\ge 1+\ds\f{4\ln 3-6\ln 2}{2\ln 2-\ln 3}=\ds\f{3\ln 3-4\ln 2}{2\ln 2-\ln 3}
\cong 1.81884\ldots $$
This is the least value for $a$ in this case.
{\bf Case 2.}
Assume that $y = z$.
Again, since the inequality is homogeneous,
assume that $y = z = 1$.
Then the inequality can be written as
$$\left(\ds\f{x+2}{3}\right)^a \left(\ds\f{2x+1}{3}\right)^\f{3-a}{2}\ge \ds\f{(x+1)^2}{4}.
\eqno(3)$$
Squaring, yields
$16(x+2)^{2a}(2x+1)^{3-a}\ge (x+1)^4 3^{3+a}$, or
$$\left[\ds\f{(x+2)^2}{3(2x+1)}\right]^a\ge \ds\f{27(x+1)^4}{16(2x+1)^3}.
\eqno(4)$$
That is,
$$a\ge \ds\f{\ln E}{\ln F},
\eqno(5)$$
where
$E=\ds\f{27(x+1)^4}{16(2x+1)^3}$,
$F=\ds\f{(x+2)^2}{3(2x+1)}$.
The last step is possible because
$\ln F\ge 0$ since $F\ge 1$,
which is equivalent to
$$(x-1)^2=(x+2)^2-3(2x+1)\ge 0.$$
So we need to compute the maximum value
$f(x)=\ds\f{\ln E}{\ln F}$,
for all $x$ except $x=1$.
(For $x=1$
we can check that any value of a actually gives equality.)
Now we prove that $f(x)$ is decreasing on $[0,+\infty )$ and hence
has its maximum value at $x = 0$, which is exactly the value we computed in
the previous case, as we have the same triple $(1,1,0)$.
First of all let us prove that $E\ge F$.
This is equivalent to
$$81(x+1)^4\ge 16(x+2)^2(2x+1)^2,$$
and follows by the AM-GM inequality for the numbers $x + 2$ and $2x + 1$
$$\ds\f{3(x+1)}{2}\ge 2\sqrt {(x+1)(2x+1)}.$$
Now we prove that $f‘(x)\le 0$, or, as $f(x)\ge 0$
(since $E\ge F$), it suffices to prove that
$\ds\f{f‘(x)}{f(x)}\le 0$.
As $f=\ds\f{\ln E}{\ln F}$,
we deduce
$\ds\f{f‘}{f}=\ds\f{(\ln E)‘}{\ln E}-\ds\f{(\ln F)‘}{\ln F}$.
Further
$\ds\f{(\ln E)‘}{\ln E}=\ds\f{E‘}{E\ln E}$,
and
$\ds\f{(\ln F)‘}{\ln F}=\ds\f{F‘}{F\ln F}$.
So we need to prove that
$$\ds\f{E‘}{E}\ln F-\ds\f{F‘}{F}\ln E\ge 0.$$
As
$$\ds\f{E‘}{E}=\ds\f{2(x-1)}{(x+1)(2x+1)},
\eqno(6)$$
$$\ds\f{F‘}{F}=\ds\f{2(x-1)}{(x+2)(2x+1)}
\eqno(7)$$
and using the fact that
$E\ge F\ge 1$ or $\ln E\ge \ln F\ge 0$
it suffices to show that
$$g(x)=(x-1)((x+2)\ln F-(x+1)\ln E)\ge 0.
\eqno(8)$$
Let $q(x)=(x+2)\ln F-(x+1)\ln E$.
Then
$$q‘(x)=((x+2)\ln F-(x+1)\ln E)‘=\ln F-\ln E+(x+2)\ds\f{F‘}{F}-(x+1)\ds\f{E‘}{E}.$$
As
$(x+2)\ds\f{F‘}{F}=(x+1)\ds\f{E‘}{E}$
because of (6) and (7) we deduce that
$$q‘(x)=\ln F-\ln E\le 0,$$
so this function is decreasing.
Let us look at
$g(x)=(x-1)q(x)$.
For a decreasing function $q(x)$ we have
$$x=1:\q
q(x)=(x+2)\ln F-(x+1)\ln E=0$$
$$x<1:\q
q(x)=(x+2)\ln F-(x+1)\ln E\ge0$$
$$x>1:\q
q(x)=(x+2)\ln F-(x+1)\ln E\le 0$$
so in each case
$g(x)=(x-1)((x+2)\ln F-(x+1)\ln E)=(x-1)q(x)\le 0$
thus $f‘(x)\le 0$
and the greatest value is at 0.
We have already proved in the first case that
$$a\ge \ds\f{3\ln 3-4\ln 2}{2\ln 2-\ln 3}=1.81884\ldots $$
and our proof is complete.
\bigskip
\hfill
{\Large Iurie Boreico and Ivan Borsenco, USA}
翻译事宜
标签：cal must keep 结果 phi $? triple can max
原文地址：http://www.cnblogs.com/Eufisky/p/7771987.html
