Maths Extension 2 � Harder Maths Extension 1 (3U)

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Harder Maths Extension 1 / Harder 3U

! Inequalities! Induction

Maths Extension 2 � Harder Maths Extension 1 (3U)

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InequalitiesProof22 2 baba +− 0≥

( )22 ba + 0≥abba 222 ≥+True ifTrue if

ba =ba ≠

Example 1PROVE ( )2ba + ab4≥

( )2ba + = 22 2 baba ++= ( ) abba 42 +−

∴ ( )2ba + ab4≥

Example 2If a, b, c, d are > 0, then abcddcba 44444 ≥+++PROVE 222 cba ++ cabcab ++≥

( ) ( ) ( )222222 accbba +++++ cabcab 222 ++≥ Using

( )2222 cba ++ ( )cabcab ++≥ abba 222 ≥+∴ 222 cba ++ cabcab ++≥

Example 3

The arithmetic mean =2ba +

The geometric mean = abPROVE

2ba +

≥ ab

( )4

2ba + ≥ ab

( )2ba + ≥ 4ab22 2 baba ++ ≥ 4ab22 2 baba +− ≥ 0

( )2ba − ≥ 0

True if ba =True if ba ≠

Maths Extension 2 � Harder Maths Extension 1 (3U)

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Example 4

Are they equal?ba +

1 = ba11 +

ba +1 ≤

ba11 +

≤abab +

ab ≤ ( )2ba +abbaba −++ 22 2 ≥ 0

if 22 baba ++ ≥ 0

Prove by contradiction

ba11 + ≥

ba +1

abab + ≥

ba +1

( )2ba + ≥ ab

Since ( )2ba + ≥ ab2( )2ba + ≥ ab

Maths Extension 2 � Harder Maths Extension 1 (3U)

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InductionExample 1

( )xf = xxe = ( )0+xex

( )xf ' = xx exe + = ( )1+xex

( )xf '' = xxx eexe ++ = ( )2+xex

( )xf ''' = = ( )3+xex

( )xf n = = ( )nxex +

Prove for n = 0LHS = ( )xf 0 RHS = ( )0+xex

= xxe = xxe

Assume n = k Assume n = k + 1( )xf k = ( )kxex + ( )xf k 1+ = ( )1++ kxex

= xx kexe + = xxx keexe ++( )xf k 1+ = xxx keexe ++

Maths Extension 2 � Harder Maths Extension 1 (3U)

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Example 2Recurrence InductionA sequence of terms nU , n = 1, 2, 3, �

nn UT =

321 254 −−− +−= nnnn UUUU n = 4, 5, �Initial conditions 31 =U , 12 =U , 03 =U

Show that by induction532 1 +−= − nU n

nn = 1, 2, 3, �

Consider S(n) = 532 1 +−− nn

S(1) = 5)1(32 11 +−−

= 1 � 3 + 5= 3

S(2) = 5)2(32 12 +−−

= 2 � 6 + 5= 1

S(3) == 4 � 9 + 5= 0

Assume S(k) = 532 1 +−− kk

Prove S(k+1) = 5)1(32 11 ++−−+ kk

= 5332 +−− kk

= 232 +− kk

LHS = 321 254 −−− +− kkk UUU RHS = 232 +− kk

=( ) ( ) ( )5)2(3225)1(3255324 321 +−−++−−−+− −−− kkk kkk

= 226240152.520122 221 −−+−+−+− −−+ kkk kk= ( ) 2322.522 22 +−+− −− kk

= 232 +− kk

Hence if n = k True for n = k + 1True for n = 1, 2, 3, �∴ must be true for all positive integer n