section{Trigonometric Substitutions with Broad Results}%32
markboth{Articles}{Trigonometric Substitutions with Broad Results}
具有广泛结果的三角代换
vspace{4.2cm}
Often the key to solve some intricate algebraic inequality is to simplify it with
a trigonometric substitution.
When we make a trigonometric substitution the
problem may reduce so much that we can see a direct solution immediately.
Besides, trigonometric functions have well-known properties that may help in
solving such inequalities.
As a result, many algebraic problems can be solved
by using an inspired substitution.
We start by introducing such substitutions.
After that we present some
well-known trigonometric identities and inequalities.
Finally, we discuss some
Olympiad problems and leave others for the reader to solve.
通常，解决一些复杂代数不等式的关键是用三角代换来简化它。当我们作三角替换时，问题可能会减少很多，我们马上就能看到一个直接的解答。此外，三角函数具有熟知的性质，它们可以帮助解决这些不等式。因此，用一个启发性的代换方法可以解决许多代数问题。
我们从引入这样的代换开始。在此之后，我们提出了一些著名的三角恒等式和不等式。最后，我们讨论了一些奥林匹克问题，并留给读者去解决。
定理1. 令$alpha ,eta ,gamma $为$(0,pi )$中的角度.那么$alpha ,eta ,gamma $是某三角形的内角当且仅当
$$ anfrac{alpha }{2} anfrac{eta }{2}+ an frac{eta }{2} an frac{gamma }{2}
+ an frac{gamma }{2} an frac{alpha }{2}=1.$$
证明.首先注意到若$alpha =eta =gamma $,则命题显然成立。不失一般性，假设$alpha e eta $.因为$0<alpha +eta <2pi $,由此可知$(-pi ,pi )$中存在某个角，记为$gamma '$, 使得$alpha +eta +gamma '=pi $.利用加法公式和这样的事实
$$ an frac{gamma '}{2}=cotfrac{alpha +eta }{2}
=frac{1- an frac{alpha }{2} an frac{eta }{2}}{ an frac{alpha }{2}+ an frac{eta }{2}},$$
可得
$$ an frac{alpha }{2} an frac{eta }{2}+ an frac{eta }{2} an frac{gamma '}{2}
+ an frac{gamma '}{2} an frac{alpha }{2}=1.
eqno(1)$$
现假设
$$ anfrac{alpha }{2} andsf{eta }{2}+ an dsf{eta }{2} an dsf{gamma }{2}
+ an dsf{gamma }{2} an dsf{alpha }{2}=1,
eqno(2)$$
对某些$(0,pi )$中的$alpha ,eta ,gamma $成立。我们将证明$gamma =gamma '$, 并且这将表明$alpha ,eta ,gamma $是某个三角形的内角。
(1) 减去 (2)可得
$ an dsf{gamma }{2}= an dsf{gamma '}{2}$.因此
$left|dsf{gamma -gamma '}{2}
ight|=kpi $
对某个非负整数$k$成立.但是
$left|dsf{gamma -gamma '}{2}
ight|le left|dsf{gamma }{2}
ight|
+left|dsf{gamma '}{2}
ight|<pi $,故我们可知$k=0$.也就是$gamma =gamma '$, 这便是所需结果。
定理2.令$alpha ,eta ,gamma $为$(0,pi )$中的角.那么$alpha ,eta ,gamma $是三角形的内角当且仅当
$$sin^2 dsf{alpha }{2}+sin^2 dsf{eta }{2}+sin^2 dsf{gamma }{2}
+2sindsf{alpha }{2}sindsf{eta }{2}sindsf{gamma }{2}=1.$$
因为$0<alpha +eta <2pi $, 在$(-pi ,pi )$中存在某个角，记为$gamma '$使得$alpha +eta +gamma '=pi $.
利用积化和差和倍角公式，我们得到
$$sin^2dsf{gamma '}{2}+2sindsf{alpha }{2}sindsf{eta }{2}sindsf{gamma '}{2}
=cosdsf{alpha +eta }{2}left(cosdsf{alpha +eta }{2}
+2sindsf{alpha }{2}sindsf{eta }{2}
ight)$$
$$ =cosdsf{alpha +eta }{2}cosdsf{alpha -eta }{2}
=dsf{cos alpha +coseta }{2}
=dsf{left(1-2sin^2 dsf{alpha }{2}
ight)+left(1-2sin^2 dsf{eta }{2}
ight)}{2}$$
$$ =1-sin^2 dsf{alpha }{2}-sin^2dsf{eta }{2}.$$
因此
$$sin^2 dsf{alpha }{2}+sin^2dsf{eta }{2}+sin^2 dsf{gamma '}{2}
+2sindsf{alpha }{2}sindsf{eta }{2}sindsf{gamma '}{2}=1.
eqno(3)$$
现假设
$$sin^2 dsf{alpha }{2}+sin^2dsf{eta }{2}+sin^2 dsf{gamma }{2}
+2sindsf{alpha }{2}sindsf{eta }{2}sindsf{gamma }{2}=1,
eqno(4)$$
对某些$(0,pi )$中的$alpha ,eta ,gamma $成立。(3)减去(4)可得
$$sin^2dsf{gamma }{2}-sin^2dsf{gamma '}{2}
+2sindsf{alpha }{2}sindsf{eta }{2}left(sindsf{gamma }{2}-sindsf{gamma '}{2}
ight)=0,$$
即
$$left(sindsf{gamma }{2}-sindsf{gamma '}{2}
ight)
left(sindsf{gamma }{2}+sindsf{gamma '}{2}+2sindsf{alpha }{2}sindsf{eta }{2}
ight)
=0.$$
第二个因式可以被写成
$$sindsf{gamma }{2}+sindsf{gamma '}{2}+cosdsf{alpha -eta }{2}
-cosdsf{alpha +eta }{2}=sindsf{gamma }{2}+cosdsf{alpha -eta }{2},$$
它是明显大于$0$的。由此得
$sindsf{gamma }{2}=sindsf{gamma '}{2}$,故$gamma =gamma '$,这就证明了$alpha ,eta ,gamma $是某个三角形的内角。
subsection{代换和转换}
转换1.令$alpha ,eta ,gamma $为三角形的内角。令$$A=dsf{pi -alpha }{2},q
B=dsf{pi -eta }{2},q
C=dsf{pi -gamma }{2}.$$那么$A+B+C=pi $, 且$0le A,B,Cle dsf{pi }{2}$.
这个变换使我们可以将任意三角形的角度转换成一个锐角三角形的角度。注意到
$$sindsf{alpha }{2}=cos A,q
cosdsf{alpha }{2}=sin A,q
andsf{alpha }{2}=cot A,q
cotdsf{alpha }{2}= an A,$$
类似的结果对$eta $和$gamma $也成立。
转换2.令$x,y,z$为正实数，那么存在一个以$a=x+y$, $b=y+z$, $c=z+x$为边长的三角形。该转换有时被称为对偶原理(Dual Principle).显然，若$s=x+y+z$,则 $(x,y,z)=(s-b,s-c,s-a)$.
代换1.令$a,b,c$为正实数，使得$ab+bc+ca=1$.利用函数
$f:left(0,dsf{pi }{2}
ight) o (0,+infty )$, 且
$f(x)= an x$,我们可作如下代换
$$a= an dsf{alpha }{2},q
b= an dsf{eta }{2},q
c= an dsf{gamma }{2},$$
其中$alpha ,eta ,gamma $为某三角形$ABC$的内角。
{f S2.}
Let $a,b,c$ be positive real numbers such that $ab+bc+ca=1$.
Applying T1 to S1, we have
$$a=cot A,q
b=cot B,q
c=cot C,$$
where $A,B,C$ are the angles of an acute triangle.
{f S3.}
Let $a,b,c$ be positive real numbers such that $a+b+c=abc$.
Dividing by $abc$ it follows that
$dsf{1}{bc}+dsf{1}{ca}+dsf{1}{ab}=1$.
Due to S1, we can substitute
$$dsf{1}{a}= andsf{alpha }{2},q
dsf{1}{b}= andsf{eta }{2},q
dsf{1}{c}= andsf{gamma }{2},$$
that is
$$a=cosdsf{alpha }{2},q
b=cotdsf{eta }{2},q
c=cosdsf{gamma }{2},$$
where $alpha ,eta ,gamma $ are the angles of a triangle.
{f S4.}
Let $a,b,c$ be positive real numbers such that $a+b+c=abc$.
Applying T1 to S3, we have
$$a= an A,q
b= an B,q
c= an C,$$
where $A,B,C$ are the angles of an acute triangle.
{f S5.}
Let $a,b,c$ be positive real numbers such that
$a^2+b^2+c^2+2abc=1$.
Note that since all the numbers are positive it follows that $a,b,c<1$.
Using the function $f:(0,pi ) o (0,1)$, with $f(x)=sindsf{x}{2}$,
and recalling Theorem 2, we can substitute
$$a=sindsf{alpha }{2},q
b=sindsf{eta }{2},q
c=sindsf{gamma }{2},$$
where $alpha,eta ,gamma $ are the angles of a triangle.
{f S6.}
Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2+2abc=1$.
Applying T1 to S5, we have
$$a=cos A,q
b=cos B,q
c=cos C,$$
where $A,B,C$ are the angles of an acute triangle.
{f S7.}
Let $x,y,z$ be positive real numbers.
Applying T2 to expressions
$$sqrt {dsf{yz}{(x+y)(x+z)}},q
sqrt {dsf{zx}{(y+z)(y+x)}},q
sqrt {dsf{xy}{(z+x)(z+y)}},$$
they can be substituted by
$$sqrt {dsf{(s-b)(s-c)}{bc}},q
sqrt {dsf{(s-c)(s-a)}{ca}},q
sqrt {dsf{(s-a)(s-b)}{ab}},$$
where $a,b,c$ are the side-lengths of a triangle.
Recall the following identities
$$sindsf{alpha }{2}=sqrt {dsf{(s-b)(s-c)}{bc}},q
cosdsf{alpha }{2}=sqrt {dsf{s(s-a)}{bc}}.$$
Thus our expressions can be substituted by
$sindsf{alpha }{2}$, $sindsf{eta }{2}$, $sindsf{gamma }{2}$,
where $alpha ,eta ,gamma $ are the angles of a triangle.
{f S8.}
Analogously to S7, the expressions
$$sqrt {dsf{x(x+y+z)}{(x+y)(x+z)}},q
sqrt {dsf{y(x+y+z)}{(y+z)(y+x)}},q
sqrt {dsf{z(x+y+z)}{(z+x)(z+y)}},$$
can be substituted by
$cosdsf{alpha }{2}$, $cosdsf{eta }{2}$, $cosdsf{gamma }{2}$,
where $alpha ,eta ,gamma $ are the angles of a triangle.
We present a list of inequalities and equalities that can be helpful in solving
many problems or simplify them.
subsection{Well-known Inequalities}
Let $alpha ,eta ,gamma $ be angles of a triangle $ABC$.
Then
medskip
1) $cos alpha +coseta +cosgamma le sindsf{alpha }{2}+sindsf{eta }{2}
+sindsf{gamma }{2}le dsf{3}{2}$
medskip
2) $sin alpha +sin eta +singamma le cosdsf{alpha }{2}+cosdsf{eta }{2}
+cosdsf{gamma }{2}le dsf{3sqrt 3}{2}$
medskip
3) $cos alpha cos eta cos gamma le sindsf{alpha }{2}sindsf{eta }{2}
sindsf{gamma }{2}le dsf{1}{8}$
medskip
4) $sinalpha sineta singamma le cosdsf{alpha }{2}cosdsf{eta }{2}
cosdsf{gamma }{2}le dsf{3sqrt 3}{8}$
medskip
5) $cot dsf{alpha }{2}+cotdsf{eta }{2}+cotdsf{gamma }{2}ge 3sqrt 3$
medskip
6) $cos^2 alpha +cos^2 eta +cos^2 gamma ge
sin^2 dsf{alpha }{2}+sin^2dsf{eta }{2}+sin^2dsf{gamma }{2}ge dsf{3}{4}$
medskip
7) $sin^2 alpha +sin^2 eta +sin^2 gamma
le cos^2dsf{alpha }{2}+cos^2dsf{eta }{2}+cos^2dsf{gamma }{2}le dsf{9}{4}$
medskip
8) $cotalpha +coteta +cotgamma ge
an dsf{alpha }{2}+ andsf{eta }{2}+ andsf{gamma }{2}ge sqrt 3$.
subsection{Well-known Identities}
Let $alpha ,eta ,gamma $ be angles of a triangle $ABC$.
Then
medskip
1) $cos alpha +cos eta +cos gamma =1+4sindsf{alpha }{2}sindsf{eta }{2}
sindsf{gamma }{2}$
medskip
2) $sinalpha +sineta +singamma =4cosdsf{alpha }{2}cosdsf{eta }{2}
cosdsf{gamma }{2}$
medskip
3) $sin2alpha +sin 2eta +sin 2gamma =4sin alpha sin eta sin gamma $
medskip
4) $sin^2 alpha +sin^2 eta +sin^2 gamma
=2+2cos alpha cos eta cos gamma $
medskip
For arbitrary angles $alpha ,eta ,gamma $ we have
$$sin alpha +sin eta +sin gamma -sin(alpha +eta +gamma )
=4sindsf{alpha +eta }{2}sindsf{eta +gamma }{2}sindsf{gamma +alpha }{2}$$
$$cos alpha +cos eta +cos gamma +cos(alpha +eta +gamma )
=4cos dsf{alpha +eta }{2}cosdsf{eta +gamma }{2}cosdsf{gamma +alpha }{2}.$$
subsection{Applications}
{f 1.}
Let $x,y,z$ be positive real numbers.
Prove that
$$dsf{x}{x+sqrt {(x+y)(x+z)}}+dsf{y}{y+sqrt {(y+z)(y+x)}}
+dsf{z}{x+sqrt {(z+x)(z+y)}}le 1.$$
hfill
(Walther Janous, Crux Mathematicorum)
{f Solution.}
The inequality is equivalent to
$$sum_{cyc}dsf{1}{1+sqrt {dsf{(x+y)(x+z)}{x^2}}}le 1.$$
Because the inequality is homogeneous, we can assume that
$xy+yz+zx=1$.
Let us apply substitution S1:
$cycleft(x= andsf{alpha }{2}
ight)$,
where $alpha ,eta ,gamma $ are angles of a triangle.
We get
$$dsf{(x+y)(x+z)}{x^2}
=dsf{left( andsf{alpha }{2}+ andsf{eta }{2}
ight)
left( andsf{alpha }{2}+ andsf{gamma }{2}
ight)}{ an^2 dsf{alpha }{2}}
=dsf{1}{sin^2 dsf{alpha }{2}},$$
and similar expressions for the other terms.
The inequality becomes
$$dsf{sindsf{alpha }{2}}{1+sindsf{alpha }{2}}
+dsf{sindsf{eta }{2}}{1+sindsf{eta }{2}}
+dsf{sindsf{gamma }{2}}{1+sindsf{gamma }{2}}le 1,$$
that is
$$2le dsf{1}{1+sindsf{alpha }{2}}+dsf{1}{1+sindsf{eta }{2}}
+dsf{1}{1+sindsf{gamma }{2}}.$$
On the other hand, using the well-known inequality
$sindsf{alpha }{2}+sindsf{eta }{2}+sindsf{gamma }{2}
le dsf{3}{2}$
and the Cauchy-Schwarz inequality, we have
$$2le dsf{9}{left(1+sindsf{alpha }{2}
ight)
+left(1+sindsf{eta }{2}
ight)+left(1+sindsf{gamma }{2}
ight)}
le sum_{cyc}dsf{1}{1sindsf{alpha }{2}},$$
and we are done.
{f 2.}
Let $x,y,z$ be real numbers greater than 1 such that
$dsf{1}{x}+dsf{1}{y}+dsf{1}{z}=2$.
Prove that
$$sqrt {x-1}+sqrt {y-1}+sqrt {z-1}le sqrt {x+y+z}.$$
hfill
(Iran, 1997)
{f Solution.}
Let $(x,y,z)=(a+1,b+1,c+1)$, with $a,b,c$ positive real numbers.
Note that the hypothesis is equivalent to $ab+bc+ca+2abc=1$.
Then it suffices to prove that
$$sqrt a+sqrt b+sqrt cle sqrt {a+b+c+3}.$$
Squaring both sides of the inequality and canceling some terms yields
$$sqrt {ab}+sqrt {bc}+sqrt {ca}le dsf{3}{2}.$$
Using substitution S5 we get
$(ab,bc,ca)=left(sin^2 dsf{alpha }{2},sin^2dsf{eta }{2},sin^2dsf{gamma }{2}
ight)$,
where $ABC$ is an arbitrary triangle.
The problem reduces to proving that
$$sindsf{alpha }{2}+sindsf{eta }{2}+sindsf{gamma }{2}le dsf{3}{2},$$
which is well-known, and can be done using Jensen inequality.
{f 3.}
Let $a,b,c$ be positive real numbers such that $a+b+c=1$.
Prove that
$$sqrt {dsf{ab}{c+ab}}+sqrt {dsf{bc}{a+bc}}+sqrt {dsf{ca}{b+ca}}
le dsf{3}{2}.$$
{f Solution.}
The inequality is equivalent to
$$sqrt {dsf{ab}{(c+a)(c+b)}}+sqrt {dsf{bc}{(a+b)(a+c)}}
+sqrt {dsf{ca}{(b+c)(b+a)}}le dsf{3}{2}.$$
Substitution S7 replaces the three terms in the inequality by
$sindsf{alpha }{2}$, $sindsf{eta }{2}$, $sindsf{gamma }{2}$.
Thus it suffices to prove
$sindsf{alpha }{2}+sindsf{eta }{2}+sindsf{gamma }{2}le dsf{3}{2}$,
which clearly holds.
{f 4.}
Let $a,b,c$ be positive real numbers such that
$(a+b)(b+c)(c+a)=1$.
Prove that
$$ab+bc+cale dsf{3}{3}.$$
hfill
(Cezar Lupu, Romania, 2005)
{f Solution.}
Observe that the inequality is equivalent to
$$left(sum_{cyc}ab
ight)^3le left(dsf{3}{4}
ight)^3 (a+b)^2(b+c)^2(c+a)^2.$$
Because the inequality is homogeneous, we can assume that
$ab+bc+ca=1$.
We use substitution S1:
$cycleft(a= andsf{alpha }{2}
ight)$,
where $alpha ,eta ,gamma $ are the angles of a triangle.
Note that
$$(a+b)(b+c)(c+a)=prod left(dsf{cosdsf{gamma }{2}}
{cosdsf{alpha }{2}cosdsf{eta }{2}}
ight)
=dsf{1}{cosdsf{alpha }{2}cosdsf{eta }{2}cosdsf{gamma }{2}}.$$
Thus it suffices to prove that
$$left(dsf{4}{3}
ight)^3le dsf{1}{cos^2 dsf{alpha }{2}cos^2 dsf{eta }{2}
cos^2 dsf{gamma }{2}},$$
or
$$4cos dsf{alpha }{2}cosdsf{eta }{2}cosdsf{gamma }{2}le dsf{3sqrt 3}{2}.$$
From the identity
$4cos dsf{alpha }{2}cosdsf{eta }{2}cosdsf{gamma }{2}=sinalpha +sineta +singamma $,
the inequality is equivalent to
$$sin alpha +sineta +singamma le dsf{3sqrt 3}{2}.$$
But $f(x)=sin x$ is a concave function on $(0,pi )$ and the conclusion follows
from Jensen's inequality.
{f 5.}
Let $a,b,c$ be positive real numbers such that $a+b+c=1$.
Prove that
$$a^2+b^2+c^2+2sqrt {3abc}le 1.$$
hfill
(Poland, 1999)
{f Solution.}
Let
$cycleft(x=sqrt {dsf{bc}{a}}
ight)$.
It follows that $cyc(a=yz)$.
The inequality becomes
$$x^2y^2+y^2z^2+z^2x^2+2sqrt 3 xyzle 1,$$
where $x,y,z$ are positive real numbers such that
$xy+yz+zx=1$.
Note that the inequality is equivalent to
$$(xy+yz+zx)^2+2sqrt 3xyzle 1+2xyz(x+y+z),$$
or
$$sqrt 3le x+y+z.$$
Applying substitution S1
$cycleft(x= andsf{alpha }{2}
ight)$,
it suffices to prove that
$$ andsf{alpha }{2}+ andsf{eta }{2}+ andsf{gamma }{2}ge sqrt 3.$$
The last inequality clearly holds, as
$f(x)= andsf{x}{2}$
is convex function on $(0,pi )$, and the conclusion follows from Jensen's inequality.
{f 6.}
Let $x,y,z$ be positive real numbers.
Prove that
$$sqrt {x(y+z)}+sqrt {y(z+x)}+sqrt {z(x+y)}
ge 2sqrt {dsf{(x+y)(y+z)(z+x)}{x+y+z}}.$$
{f Solution.}
Rewrite the inequality as
$$sqrt {dsf{x(x+y+z)}{(x+y)(x+z)}}+sqrt {dsf{y(x+y+z)}{(y+z)(y+x)}}
+sqrt {dsf{z(x+y+z)}{(z+x)(z+y)}}ge 2.$$
Applying substitution S8, it suffices to prove that
$$cosdsf{alpha }{2}+cosdsf{eta }{2}+cosdsf{gamma }{2}ge 2,$$
where $alpha ,eta ,gamma $ are the angles of a triangle.
Using transformation T1
$cycleft(A=dsf{pi -alpha }{2}
ight)$,
where $A,B,C$ are angles of an acute triangle, the inequality is equivalent to
$$sin A+sin B+sin Cge 2.$$
There are many ways to prove this fact.
We prefer to use Jordan's inequality, that is
$$dsf{2alpha }{pi }le sin alpha le alpha
mbox{ for all }
alpha in left(0,dsf{pi }{2}
ight).$$
The conclusion immediately follows.
{f 7.}
Let $a,b,cin (0,1)$ be real numbers such that $ab+bc+ca=1$.
Prove that
$$dsf{a}{1-a^2}+dsf{b}{1-b^2}+dsf{c}{1-c^2}
ge dsf{3}{4}left(dsf{1-a^2}{a}+dsf{1-b^2}{b}+dsf{1-c^2}{c}
ight).$$
hfill
(Calin Popa)
{f Solution.}
We apply substitution S1 $cycleft(a= andsf{A}{2}
ight)$, where $A,B,C$
are angles of a triangle.
Because $a,b,cin (0,1)$, it follows that
$ andsf{A}{2}, andsf{B}{2}, an{C}{2}in (0,1)$,
that is $A,B,C$ are angles of an acute triangle.
Note that
$$cycleft(dsf{a}{1-a^2}=dsf{sindsf{A}{2}cosdsf{A}{2}}{cos A}
=dsf{ an A}{2}
ight).$$
Thus the inequality is equivalent to
$$ an A+ an B+ an Cge 3left(dsf{1}{ an A}+dsf{1}{ an B}+dsf{1}{ an C}
ight).$$
Now observe that if we apply transformation T1 and the result in Theorem 1, we get
$$ an A+ an B+ an C= an A an B an C.$$
Hence our inequality is equivalent to
$$( an A+ an B+ an C)^2ge 3( an A an B+ an B an C+ an A an C).$$
This can be written as
$$dsf{1}{2}( an A- an B)^2+( an B- an C)^2+( an C- an A)^2ge 0,$$
and we are done.
{f 8.}
Let $x,y,z$ be positive real numbers.
Prove that
$$sqrt {dsf{y+z}{x}}+sqrt {dsf{z+x}{y}}+sqrt {dsf{x+y}{z}}
ge sqrt {dsf{16(x+y+z)^3}{3(x+y)(y+z)(z+x)}}.$$
hfill
Vo Quoc Ba Can, Mathematical Reflections, 2007
{f Solution.}
Note that the inequality is equivalent to
$$sum_{cyc}(y+z)sqrt {dsf{(x+y)(z+x)}{x(x+y+z)}}ge dsf{4(x+y+z)}{sqrt 3}.$$
Let use transformation T2 and substitution S8.
We get
$$cycleft((y+z)sqrt {dsf{(x+y)(z+x)}{x(x+y+z)}}=dsf{a}{cosdsf{alpha }{2}}
=4Rsindsf{alpha }{2}
ight),$$
and
$$dsf{4(x+y+z)}{sqrt 3}=dsf{4R(sinalpha +sineta +singamma )}{sqrt 3},$$
where $alpha ,eta ,gamma $ are angles of a triangle with circumradius $R$.
Therefore it suffices to prove that
$$dsf{sqrt 3}{2}left(sindsf{alpha }{2}+sindsf{eta }{2}+sindsf{gamma }{2}
ight)
ge sindsf{alpha }{2}cosdsf{alpha }{2}+sindsf{eta }{2}cosdsf{eta }{2}
+sindsf{gamma }{2}cosdsf{gamma }{2}.$$
Because $f(x)=cosdsf{x}{2}$
is a concave function on $[0,pi ]$, from Jensen's inequality we obtain
$$dsf{sqrt 3}{2}ge dsf{1}{3}left(cosdsf{alpha }{2}+cosdsf{eta }{2}
+cosdsf{gamma }{2}
ight).$$
Finally, we observe that $f(x)=sindsf{x}{2}$ is an increasing function on $[0,pi ]$,
while $g(x)=cosdsf{x}{2}$ is a decreasing function on $[0,pi ]$.
Using Chebyshev's inequality, we have
$$dsf{1}{3}left(sindsf{alpha }{2}+sindsf{eta }{2}+sindsf{gamma }{2}
ight)
left(cosdsf{alpha }{2}+cosdsf{eta }{2}+cosdsf{gamma }{2}
ight)
ge sum sindsf{alpha }{2}cosdsf{alpha }{2},$$
and the conclusion follows.
subsection{Problems for Independent Study}
{f 1.}
Let $a,b,c$ be positive real numbers such that $a+b+c=1$.
Prove that
$$dsf{a}{sqrt {b+c}}+dsf{b}{sqrt {c+a}}+dsf{c}{sqrt {a+b}}ge sqrt {dsf{3}{2}}.$$
hfill
(Romanian Mathematical Olympiad, 2005)
{f 2.}
Let $a,b,c$ be positive real numbers such that $a+b+c=1$.
Prove that
$$sqrt {dsf{1}{a}-1}sqrt {dsf{1}{b}-1}+sqrt {dsf{1}{b}-1}sqrt {dsf{1}{c}-1}
+sqrt {dsf{1}{c}-1}sqrt {dsf{1}{a}-1}ge 6.$$
{f 3.}
Let $a,b,c$ be positive real numbers such that $ab+bc+ca=1$.
Prove that
$$dsf{1}{sqrt {a+b}}+dsf{1}{sqrt {b+c}}+dsf{1}{sqrt {c+a}}ge 2+dsf{1}{sqrt 2}.$$
hfill
(Le Trung Kien)
{f 4.}
Prove that for all positive real numbers $a,b,c$,
$$(a^2+2)(b^2+2)(c^2+2)ge 9(ab+bc+ca).$$
hfill
(APMO, 2004)
{f 5.}
Let $x,y,z$ be positive real numbers such that $dsf{1}{x}+dsf{1}{y}+dsf{1}{z}=1$.
Prove that
$$sqrt {x+yz}+sqrt {x+yz}+sqrt {x+yz}ge sqrt {xyz}+sqrt x+sqrt y+sqrt z.$$
hfill
(APMO, 2002)
{f 6.}
Let $a,b,c$ be positive real numbers.
Prove that
$$dsf{b+c}{a}+dsf{c+a}{b}+dsf{a+b}{c}
ge 4left(dsf{a}{b+c}+dsf{b}{c+a}+dsf{c}{a+b}
ight).$$
hfill
(Mircea Lascu)
{f 7.}
Let $a,b,c$ be positive real numbers, such that
$a+b+c=sqrt {abc}$.
Prove that
$$ab+bc+cage 9(a+b+c).$$
hfill
(Belarus, 1996)
{f 8.}
Let $a,b,c$ be positive real numbers such that $a+b+c=abc$.
Prove that
$$(a-1)(b-1)(c-1)le 6sqrt 3-10.$$
hfill
(Gabriel Dospinescu, Marian Tetiva)
{f 9.}
Let $a,b,c$ be nonnegative real numbers such that
$a^2+b^2+c^2+abc=4$.
Prove that
$$0le ab+bc+ca-abcle 2.$$
hfill
(Titu Andreescu, USAMO, 2001)
section*{Bibliography}
i
item[{[1]}]
T. Andreescu, Z. Feng,
{it 103 Trigonometry Problems: From the Training of the USA IMO Team},
Birkhauser, 2004.
item[{[2]}]
T. Andreescu, V. Cartoaje, G. Dospinescu, M. Lascu,
{it Old and New Inequalities},
GIL Publishing House, 2004.
item[{[3]}]
T. Phuong,
{it Diamonds in Mathematical Inequalities},
Hanoi Publishing House, 2007.
item[{[4]}]
N.M. Sedrakyan
{it Geometricheskie Neravenstva},
Yerevan, 2004.
item[{[5]}]
E. Specht,
{it Collected Inequalities},
http://www.imo.org.yu/othercomp/Journ/ineq.pdf
item[{[6]}]
H. Lee,
{it Topics in Inequalities - Theorems and Techniques}.
item[{[7]}]
H. Lee,
{it Inequalities through problems}.
item[{[8]}]
Crux Mathematicorum and Mathematical Mayhem (Canada).
ei
igskip
hfill
{Large Vardan Verdiyan, Yerevan, Armenia}
hfill
{Large Daniel Campos Salas, Costa Rica}
%%%%%%%
ewpage
section{A Factoring Lemma}%33
markboth{Articles}{A Factoring Lemma}
vspace{4.2cm}
subsection{Introduction}
The following classical problem was proposed at the Moldavian Mathematical
Olympiad.
{f Problem 1.}
{it Let $a,b,c,d$ be positive integers with $ab = cd$.
Prove that $a+b+c + d$ is composite.
}
{f Solution.}
Let $N = a + b + c + d$.
Then
$$aN = a^2 + ab + ac + ad.
eqno(1)$$
Because to $ab = cd$, we can rewrite (1) as
$$aN = a^2 + cd + ac + ad = (a + c)(a + d).$$
Both $a + c$ and $a + d$ are greater than $a$, so each contributes a factor greater
than 1 to $N$.
Thus $N$ is composite.
This solution relies on a clever algebraic trick.
However, there is a more
consistent and motivated solution that relies on the Factoring Lemma.
{f Lemma 1.}
{it Let $a,b,c,d$ be positive integers with $ab = cd$.
Then there are positive integers $m,n,p,q$ such that $gcd(n,p)=1$ and
$$a=mn,q
b=pq,q
c=mp,q
d=nq.$$
}
{f Proof.}
The condition $ab = cd$ can be written as
$$dsf{a}{c}=dsf{d}{b}.$$
Let both fractions have the same representation $dsf{n}{p}$
in lowest terms.
Then
$$m=dsf{a}{n}=dsf{c}{p},q
q=dsf{b}{p}=dsf{d}{n}.$$
It is clear that $m$ and $q$ are integers and $m,n,p,q$ have the desired properties.
{f Second Solution to Problem 1.}
We can find positive integers $m,n,p,q$
satisfying the conditions in Lemma 1.
Then
$$a + b + c + d = mn + pq + mp + nq = (m + n)(p + q),$$
which explicitly proves that $a + b + c + d$ is composite.
Not only does Lemma 1 make problem 25 trivial, but it also appears to be
useful in many other problems proposed to national and international mathematical
olympiads.
The techniques can be applied in solving equations and
systems of equations over integers.
Interesting results can be obtained when
this lemma is used in connection with Euclidean rings other than the integers.
subsection{Diophantine Equations}
The problem of Pythagorean triples can be solved using the Factoring Lemma.
{f Problem 2.}
{it Pairwise relatively prime integers $a,b,c$ satisfy
$a^2+b^2=c^2$.
If $a$ is odd, then there exist integers $u$ and $v$ such that
$a=u^2-v^2$ and $b=2uv$.
}
{f Solution.}
Let us write the given condition as
$$b^2=(c-a)(c+a).$$
By Lemma 1 there are $m,n,p,q$ such that
$b=mn=pq$, $c-a=mp$, $c+a=nq$.
Again, by Lemma 1, there are
$x,y,z,w$ such that
$m=xy$, $n=zw$, $p=xz$, $q=yw$.
This means that $b = xyzw$ and
$$a=dsf{nq-mp}{2}=dsf{yz}{2}(w^2-x^2).$$
Because $w^2$ and $x^2$ are congruent to either 0 or 1 modulo 4 and $a$ is odd, it
follows that
$2mid yz$.
On the other hand, $yz$ divides both $b$ and $2a$, so $yz = 2$.
This finishes the proof.
{f Problem 3.}
{it Pairwise relatively prime positive integers $a,b,c$ satisfy
$a^2+b^2=2c^2$.
Prove that there are integers $u$ and $v$ such that
egin{align*}
& a=u^2-v^2+2uv\
& b=v^2-u^2pm 2uv\
& c=u^2+v^2.
end{align*}
}
{f Solution.}
We have
$$(a - c)(a + c) = (c - b)(c + b).$$
By Lemma 1, there are $m,n,p,q$ such that
$$a-c=mn,q
a+c=pq,q
c-b=mp,q
c+b=nq.$$
We thus have
$$pq - mn = mp + nq,$$
implying
$$p(q - m) = n(q + m).$$
Again, there are $x,y,z,w$ such that
$$p=xy,q
q-m=zw,q
n=xz,q
q+m=yw.$$
Then
$$a=dsf{1}{2}(mn+pq)=dsf{1}{2}left(dsf{yw-zw}{2}xz+dsf{zw+yw}{2}xy
ight)
=dsf{xw}{4}(y^2+2zy-z^2)$$
and
$$b=dsf{1}{2}(nq-mp)=dsf{1}{2}left(dsf{yw+zw}{2}xz-dsf{yw-zw}{2}xy
ight)
=dsf{xw}{4}(-y^2+2zy+z^2).$$
It follows that either $xw = 4$ and one of $z$ and $y$ is odd or $xw = 1$ and both $z$
and $y$ are even.
We then set $u = y$, $v = z$ or $u=dsf{y}{2}$, $v=dsf{z}{2}$,
respectively.
{f Problem 4.}
{it Prove that there are no positive integers $x,y,z$ such that
$$2(x^4-y^4)=z^2.$$
}
{f Solution.}
Suppose there are such $x,y,z$.
Then consider such a triple with the minimum possible $x + y + z$.
If two of $x,y,z$ are divisible by a prime $r$, then
the third number is also divisible by $r$.
If
$x=rx_1$, $y=ry_1$, $z=rz_1$, then
$$2r^2(x_1^4-y_1^4)=z_1^2.$$
Hence
$rmid z_1$ and $z_1=rz_2$, so
$$2(x_1^4-y_1^4)=z_2^2.$$
Then $x_1,y_1,z_2$ is a smaller triple.
Thus $x,y,z$ must be pairwise relatively prime.
We have
$gcd(x^2-y^2,x^2+y^2)mid gcd(2x^2,2y^2)=2$.
If it equals 1, then both $x^2-y^2$ and $x^2+y^2$ are squares, which is impossible.
Therefore
$gcd(x^2-y^2,x^2+y^2)=2$
and both $x,y$ are odd.
Thus
$$x^2-y^2=4u^2
qmbox{and}q
x^2+y^2=2v^2
eqno(2)$$
for some integers $u$ and $v$.
By Problem 2, there are integers $a$ and $b$ such that
$$u=ab,q
y=a^2-b^2,q
x=a^2+b^2.$$
Then
$$x^2=2u^2+v^2
qmbox{and}q
y^2=v^2-2u^2.$$
Thus
$(x-v)(x+v)=2u^2$ and $(v-y)(v+y)=2u^2$.
There are integers $a,b,c,d$ with
${x-v,x+v}={2a^2,4b^2}$ and ${v-y,v+y}={2c^2,4d^2}$.
Because
$u=ab=cd$, there are $m,n,p,qin mathbb{Z}$ with $a=mn$, $b=pq$, $c=mp$, $d=nq$.
Therefore $v=|a^2-2b^2|=c^2+2d^2$.
There are two possible cases.
(a) If $a^2-2b^2=c^2+2d^2$, then
$$m^2n^2=2p^2q^2+m^2p^2+2n^2q^2,q
m^2(n^2-p^2)=2q^2(n^2+p^2),$$
implying that $|n|$, $|p|$, $|mq|$ form a smaller solution.
(b) If $2b^2-a^2=c^2+2d^2$, then
$$2p^2q^2=m^2n^2+m^2p^2+2n^2q^2,q
2q^2(p^2-n^2)=m^2(n^2+p^2),$$
implying that $|n|$, $|p|$, $|mq|$ form a smaller triple.
Because both cases lead to a contradiction, the original equation has no solutions.
subsection{Sums of Two Squares and $mathbb{Z}[i]$}
The following is a well-known fact in Number Theory which also appeared as
a problem at the Moldavian IMO team selection tests in 2004.
{f Problem 5.}
{it Let $(a,b)$ and $(c,d)$ be two different unordered pairs of nonnegative
integers such that
$a^2+b^2=c^2+d^2=k$.
Prove that $k$ is composite.
}
{f Solution.}
Without loss of generality assume that $a > c$ (for if $a = c$, then
$b = d$ and the pairs would be the same).
We can rearrange the terms to obtain
$$(a - c)(a + c) = (d - b)(d + b).
eqno(3)$$
We immediately see that $d > b$.
Then $a+c$, $a-c$, $d+b$, $d-b$ are positive integers and due to (3) and Lemma 1
we can find positive integers $m,n,p,q$ with
$gcd(n,p)=1$ such that
$$a+c=mn,q
a-c=pq,q
d+b=mp,q
d-b=nq.$$
Then
$$a=dsf{mn+pq}{2},q
b=dsf{nq-mp}{2}$$
and
$$4k=4(a^2+b^2)=(mn+pq)^2+(nq-mp)^2$$
$$=m^2n^2+p^2q^2+n^2q^2+m^2p^2=(m^2+q^2)(n^2+p^2).
eqno(4)$$
Suppose $k$ is prime.
Then from (4), without loss of generality, $kmid m^2+q^2$.
Hence either
$n^2+p^2=4$ or $n^2+p^2=2$.
The former is impossible, because 4 cannot be represented as the sum of two positive
perfect squares.
The latter case yields $n = p = 1$, which implies $a = c$, $b = d$ that is again impossible.
Thus our assumption is wrong and $k$ is composite.
Note that along the way we proved the following simple statement.
{f Lemma 2.}
{it If $a,b,c,din mathbb{Z}$ and $a^2+b^2=c^2+d^2$, then there are
$m,n,p,qin mathbb{Z}$ such that
$$2a=mn+pq,q
2b=mp-nq,q
2c=mp+nq,q
2s=mn-pq.$$
}
The Factoring Lemma also holds true in the Euclidean ring $mathbb{Z}[i]$.
One application is the following lemma which leads to many interesting results.
{f Lemma 3.}
{it If $a,b,c,d$ are positive integers such that $a^2+b^2=cd$ then there
are integers $x,y,z,w,t$ such that
$$c=t(x^2+y^2),q
d=t(z^2+w^2),q
a=t(xz-yw),q
b=t(xw+yz).$$
}
{f Proof.}
Let $t=gcd(a,b,c,d)$, $a=ta_1$, $b=tb_1$, $c=tc_1$, $d=td_1$.
Then
$$a_1^2+b_1^2=c_1d_1,$$
which can be rewritten as
$$(a_1+b_1i)(a_1-b_1i)=c_1d_1.
eqno(5)$$
By Lemma 1, there are $m,n,p,qin mathbb{Z}[i]$ such that
$$a_1+b_1i=mn,q
a_1-b_1i=pq,q
c_1=np,q
d_1=mq.
eqno(6)$$
Because $np$ and $mq$ are positive integers, it follows that
$n=kov{p}$ and $q=lov{m}$
for some positive rational numbers $k$ and $l$.
On the other hand,
$|mn|=|pq|$ implies
$|kmov{p}|=|lpov{m}|$ and $k=l$.
Let $u,v$
be relatively prime positive integers such that $k=dsf{u}{v}$.
Then
$$a_1+b_1i=dsf{u}{v}mov{p},q
a_1-b_1i=dsf{u}{v}pov{m},q
c_1=dsf{u}{v}pov{p},q
d_1=dsf{u}{v}mov{m}.$$
This means that $umid a,b,c,d$ and thus $u=1$.
In addition,
$$a_1+b_1i=dsf{v}{u}nov{q},q
a_1-b_1i=dsf{v}{u}qov{n},q
c_1=dsf{v}{u}nov{n},q
d_1=dsf{v}{u}qov{q},$$
implying $vmid a,b,c,d$ and thus $v=1$.
Let $n=x+yi$ and $m=z+wi$, where $x,y,z,win mathbb{Z}$.
Then (6) implies
$$a_1=xz-yw,q
b_1=xw+yz,q
c_1=x^2+y^2,q
d_1=z^2+w^2,$$
and thus
$$a=t(xz-yw),q
b=t(xw+yz),q
c=t(x^2+y^2),q
d=t(z^2+w^2).$$
{f Corollary 1.}
{it If $a,b,c$ are positive integers such that $ab=c^2+1$, then $a$ and
$b$ can be represented as sums of two perfect squares.
}
{f Proof.}
By Lemma 2 there are integers $x,y,z,t$ such that
$$t(x^2+y^2)=a,q
t(z^2+w^2)=b,q
t(xz-yw)=c,q
t(xw+yz)=1.$$
This implies $t=1$ and $a=x^2+y^2$, $b=z^2+w^2$.
{f Corollary 2.}
{it Every prime $p$ of the form $4k + 1$ can be represented as a sum
of two integer squares.
}
{f Proof.}
By Wilson's theorem,
$$-1equiv (p-1)!=1cdot 2ldots 4k
equiv 1cdot 2ldots 2kldots (2k+1-p)(2k+2-p)ldots (4k-p)$$
$$equiv (-1)^{2k}(2k!)^2equiv (2k!)^2pmod p.
eqno(7)$$
By (7), we have
$$(2k!)^2+1=ap,$$
for some integer $a$ and by Corollary 1, $p$ is a sum of two integer squares.
subsection{Exercises}
The following problems may help the readers sharpen their skills in applying
techniques associated with the Factoring Lemma.
{f Problem 6.}
{it Find all integer solutions to the equation $x^2+3y^2=z^2$.
}
{f Problem 7.}
{it Find all integer solutions to the equation $x^2+y^2=5z^2$.
}
{f Problem 8.}
{it Prove that if $N=a^2+2b^2=c^2+2d^2$ and ${a,b}
e {c,d}$, then
$N$ is composite.
}
{f Problem 9.}
{it Prove that no four perfect squares form a non-constant arithmetic
sequence.
}
{f Problem 10.}
(IMO Shortlist, 1998)
{it Find all positive integers $n$ for which there is an integer $m$ with
$2^n-1mid m^2+9$.
}
section*{Bibliography}
i
item[{[1]}]
T. Andreescu, D.,
{it An Introduction to Diophantine Equations},
GIL Publishing House, 2003.
item[{[2]}]
T. Andreescu, D. Andrica, Z. Feng,
{it 104 Number Theory Problems: From the training of USA IMO Team},
Birkhauser, 2006.
item[{[3]}]
L. Panaitopol, A. Gica,
{it O introducere in aritmetica si teoria numerelor},
Ed. Univ. Bucuresti, 2001.
ei
igskip
hfill
{Large Iurie Boreico, USA}
hfill
{Large Roman Teleuca, Chisinau, Moldova}
%%%%%%%