using tangent lines to prove inequalities
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Transcript of using tangent lines to prove inequalities
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Phng php tip tuyn trong chng minh Bt ng thc
MC LC
Trang
1. T VN .................................................................................................22. GII QUYT VN ..................................................................................2
2.1. Nhng vn l lun !hung.....................................................................2
2.1.1. "#$ %& '()n $h*! !+n nh,..................................................................2
2.1.2. T-nh !h$............................................................................................
2.1.. N#( /ung 0hng 0h30......................................................................
2.2. Th4! $r5ng !6a vn ..............................................................................72.. 83! 9(:n 0h30 ; $()n h( ?u@)$ vn .....................................A
2.7. B(:u ?u> !6a %3ng '()n '(nh ngh(:C......................................................12
. DT EUFN...................................................................................................1
.1. TC lH! nhng g(>( 0h30.....................................................................1
.2. h5C v( 30 /Jng.....................................................................................1
Ngi thc hin: Trn Qung Th!nh 1 T"PT #$ng "uy Tr
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Phng php tip tuyn trong chng minh Bt ng thc
TI:
PHNG PHP TIP TUYN
TRONG CHNG MINH BT NG THC
----------------------------------
1. T N .
K$ Lng $h*! l< C#$ vn ?uan $rMng !6a !hng $rnh $O3n $rung hM!
0hP $hng. TrOng nh(u nRC $rS l5( @ 9$ Lng $h*! hOW! !3! 9( nh(u ^C v_n 'hng $h= h(=u H! v %aO ng]( $a l5( g(>(
H! nh v@` TrOng $
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Phng php tip tuyn trong chng minh Bt ng thc
N)u h
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Phng php tip tuyn trong chng minh Bt ng thc
Ki V()$ 0hng $rnh $()0 $u@)n !6a c $hX h( nh(u 9
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Phng php tip tuyn trong chng minh Bt ng thc
Lng $h*! \>@ ra 'h(1
.
a c= = = {
B@( +A;* !.3ho a 45 th6c a, , c; 3hng minh r:ng
2 2 2
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c a$a !
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Phng php tip tuyn trong chng minh Bt ng thc
2 2 21 Z 1 Z 1
z
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Ph'n tch"
Kmi V) $r3( !6a 9$ Lng $h*! ! !3! 9()n H! $3!h r(fng S $png h5ng $ l@ ra 'h(1
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= = =x y {
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Z 2 . Z Z Z+ + + ++ c a c a a c ac
Ph'n tch"
Kmi 8h(a ha( v) !hO ac 9$ Lng $h*! H! v()$ l5(
2.
++
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c a a
ca a
c
c
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.
+= + =
+ c
c c c a
K1i V va( $re !6a a, , cnh nhau nfn $a /4 O3n Lng $h*! \>@ ra 'h(
1= = =a c .
K2i V a, , cl< !3! %& /ng ! $Png 9ng nfn m .
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Phng php tip tuyn trong chng minh Bt ng thc
Th^O KT 8au!h@ $a !1 1 7 2
.
+= + =
+ c
c c c a
V,( CM( ( )m a $a !h*ng C(nh2 1
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+
aa
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{
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3ho a 45 th6c 7ng a, , c; 3hng minh r:ng
2 2 2 2 2 2
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+ + ++ +
+++ + ++a c c a c a
c c a a ca
Ph'n tch"
Kmi V) $r3( !6a 9$ Lng $h*! ! !3! 9()n !ha H! $3!h r(fng S $png h5ng
$ nhng 9$ Lng $h*! l< $hu+n nh$ nfn $a ! $h= !hun ha 9ng !3!h g(> %
.+ + =a c E! 9$ Lng $h*! ; !hO $rS $h@ ra 'h(
1= = =a c /O $Png !6a !hng 9ng Z.
Ngi thc hin: Trn Qung Th!nh 9 T"PT #$ng "uy Tr
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Phng php tip tuyn trong chng minh Bt ng thc
K2i (u '(:n a, , cl< !3! %& /ng ! $Png 9ng g(0 !hng $a %u@ ra
rng m .<
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Phng php tip tuyn trong chng minh Bt ng thc
2 2 2 2 2 2
Z Z Z z Z
Z Z .
AZ 2A
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a a c c a c
c caa{
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V(:! !hun ha n@ ra 'h(
m= = =a c .
K2i xt$ h
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Phng php tip tuyn trong chng minh Bt ng thc
Ki hng $rnh $()0 $u@)n !6a c $hX h?JNG'IN'INHNGHIM
3ng '()n '(nh ngh(:C nhCi
G(0 hM! %(nh $h
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Phng php tip tuyn trong chng minh Bt ng thc
G(0 hM! %(nh 9()$ !3!h vn /Jng !3! '()n $h*! $O3n C#$ !3!h l(nh hO5$
v< %3ng $5O.
n lu@:n !hO hM! %(nh ' nRng !h*ng C(nh 9$ Lng $h*! v< $C g(3 $rX
nh nh$ g(3 $rX l,n nh$.
En $hn !ng ; ! nh(u !& gbng %Ong $rOng 0h+n $rnh 9O $-!h l@ !hu@fn Cn !6a Cnh.
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Phng php tip tuyn trong chng minh Bt ng thc
TI LIU TH?M 'H"O
1. D(nY(n E( 2mm|Z K4ing tangLnt inL4 to pro8L inLNuaitiL4, "a$h^Ca$(!al
\!al(9ur 1mAZ.2. Oc -./Q-.1;
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Phng php tip tuyn trong chng minh Bt ng thc
NH&N T NH GI P LO=I C>?
T CHUYN MQN
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Cnh 'hng %aO !ht0 n#( /ung !6a
ng]( 'h3!.
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HXI NG 'H-J' C>? N Z
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CH> TZCH H 'H-J' C>? N Z
NH&N T NH GI P LO=I
C>? HXI NG 'H-J' NGNHG[T
NBFN xTi
I"i
x EIi
CH> TZCH H 'H-J' NGNH
G[T
Ngi thc hin: Trn Qung Th!nh 15 T"PT #$ng "uy Tr