Inequalities (Unit 1 – Unit 3) Solutions to Exercises Inequalities (Unit ...

4. Let a2004 = 1 − a1− a2− − a2003. Then a1 + a2 + + a2004 = 1 andaa 1 2a2003(1 −a1 −a2 − −a2003)aa1 2a2004= .( a + a + + a )(1 −a )(1 −a ) (1 −a ) (1 −a )(1 −a ) (1 −a)1 2 2003 1 2 2003 1 2 2004By the AM-GM inequality,(1 −a1 )(1 −a2) (1 −a2004)= ( a + a + + a )( a + a + + a ) ( a + a + +a )2 3 2004 1 3 2004 1 2 2003( 20032003aa2003 20032 3a2004 )( 2003 aa1 3a2004 ) ( 2003 aa1 2a2003)≥ ⋅ ⋅ ⋅= 20032004aaa1 2 2004aa1 2a20041Hence we have ≤(1 −a )(1 −a ) (1 −a) 20031 2 2004Furthermore, equality holds when1Therefore, the answer is2004200320041a1 = a2= = a2004= .2004..5. By the Cauchy-Schwarz inequality, ⎜k⎟ ⋅ ( a + b )n∑k=12⎛ ⎞nna⎛⎜ ⎟ ∑ k k≥ ⎜∑a⎝ ak+ bk⎠ k=1 ⎝ k=1k2⎞⎟⎠.Hencen∑k=12⎛ a⎜k⎝ ak+ bk⎞⎟≥⎠⎛⎜⎝∑k=1n∑k = 1n⎞⎟⎠( a + b )kak2kn⎛⎜∑ak⎝ k=1=n2 ⋅ a∑k=1⎞⎟⎠k2=n∑k = 1⎛ ak⎜⎝ 2⎞⎟ .⎠Alternative Solution2For real numbers a and b, we have ( ) 2( a + b) ≥ 2 ab = 4ab, soab a + b≤ . Hencea + b 46