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Eufisky posted @ 2017年9月24日 02:27 in 数学分析 with tags 高校资源 , 733 阅读

60和66再检查！！！

$$\frac{f\left( c^nx \right)}{c^n\left| x \right|}=\frac{cf\left( c^{n-1}x \right)}{c^n\left| x \right|}=\frac{f\left( c^{n-1}x \right)}{c^{n-1}\left| x \right|}=\cdots =\frac{f\left( x \right)}{\left| x \right|}$$
$$\frac{f\left( x \right)}{\left| x \right|}=\frac{f\left( x/c^n \right)}{\left| x \right|/c^n},$$

$$\frac{l_2}{r_2}\leq \frac{f\left( x \right)}{\left| x \right|}=\frac{f\left( c^nx \right)}{c^n\left| x \right|}\leq \frac{l_1}{r_1},\quad x\neq 0$$

$$\frac{f\left( x \right)}{\left| x \right|}=\frac{f\left( x/c^n \right)}{\left| x \right|/c^n}$$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}

$$\left(\ds\f{x+y+z}{3}\right)^3\ge \ds\f{(x+y)(y+z)(z+x)}{8}.$$

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.

{\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}

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