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`Solution of 8. of Bartle, The Elements of Real Analysis, 2/eWritten by Men-Gen Tsai email: [email protected] 8.. In this project, let {a1 , a2 , ..., an }, and so forth, be sets of n positive real numbers. (a) In can be proved (for example, by using the Mean Value Theorem) that if a and b are positive and 0 &lt;  &lt; 1, then a b1-  a + (1 - )b and that the equality holds if and only if a = b. Assume this, let r &gt; 1 and let s satisfy 1 1 + = 1, r s (so that s &gt; 1 and r + s = rs). Show that if A and B are positive, then Ar B s + , r s and that the equality holds if and only if Ar = B s . AB  (b) Let {a1 , ..., an } and {b1 , ..., bn } be positive real numbers. If r, s &gt; 1 and (1/r) + (1/s) = 1, establish H¨lder's Inequality on n 1/r n 1/raj b j j=1 j=1ar jbr jj=1(Hint: Let A = ( bj /B.)ar )1/r and B = ( jbr )1/r and apply part (a) to aj /A and j(c) Using H¨lder's Inequality, establish Minkowski Inequality on(aj + bj )rj=11/rnj=1ar j 11/rn+j=1br j1/r(Hint: (a + b)r = (a + b)(a + b)r/s = a(a + b)r/s + b(a + b)r/s .) (d) Using H¨lder's Inequality, prove that o 1 n 1 n r aj  a n j=1 n j=1 j1/r(e) If a1  a2 and b1  b2 , then (a1 - a2 )(b1 - b2 )  0 and hence a1 b 1 + a2 b 2  a1 b 2 + a2 b 1 . Show that if a1  a2  ...  an and b1  b2  ...  bn , thenn n nnj=1aj b j j=1ajj=1bj(f) Suppose that 0  a1  a2  ...  an and 0  b1  b2  ...  bn and r  1. Establish the Chebyshev Inequality 1 n r a n j=1 j1/r1 n r b n j=1 j1/r1 n (aj bj )r n j=11/rShow that this inequality must be reversed of {aj } is increasing and {bj } is decreasing. Proof of (a): Let f (x) = x + (1 - ) - x , where x = a/b. Thus f (x) =  - x-1 = (1 - x-1 ). Hence f (x) attains its minimum at x = 1 (, a = b,) and f (1) = 0. Hence f (x) = x + (1 - ) - x  0. Put x = a/b I have a a ( ) - (1 - ) - ( )  0 b b 2or a b1-  a + (1 - )b. Next, let  = 1/r, a = A1/ , and b = B 1/(1-) . Hence AB  Ar B s + . r sProof of (b): Let A = ( aj /A and bj /B:ar )1/r , B = ( jrbr )1/r and apply part (a) to jsaj b j 1 aj  AB r A for all 1  j  n. Sum them up: aj b j 1 n aj  r j=1 A j=1 A B 1 AB Thus,n n n n+1 bj s Br+1 n bj s j=1 Bsaj bj  1.j=1 1/r n 1/raj b j j=1 j=1ar jbr jj=1Proof of (c): By using H¨lder's Inequality: or/saj aj + b jr/s ar j br j1/r(aj + bj )r (aj + bj )r1/s1/r1/sb j aj + b j Thusr.aj + b j (aj + bj )r1/r (aj + bj )r ar j1/r1/sar j br j1/r1/r+br j1/r+ 3.Proof of (d): Let bj = 1/n for all j between 1 and n. Thus 1 aj  n j=1 Hencen nar jj=11/r1 ( )r j=1 n1/rn1/s.1 n 1 n r aj  a n j=1 n j=1 jProof of (e): Note that (ai - aj )(bi - bj )  0 for all i and j. Hence ai b i + aj b j  ai b j + aj b i for all i and j. Sum them up as following:n n n nai bi + aj bj j=1 i=1 n nai b j j=1 i=1 n n (bjj=1 n i=1+ aj b in(naj bj +j=1 n i=1 nai b i )  ai bi ) i=1 nai + aji=1 nbi )n nnj=1(aj bj + naii=1 n j=1bj +j=1 naji=1biHence naj b j j=1 j=1ajj=1bjNote: If a1  a2  ...  an and b1  b2  ...  bn (or a1  a2  ...  an and b1  b2  ...  bn ), thenn n nnj=1aj b j j=1ajj=1bj4It will be used in part (f). Proof of (f ): Note that 0  ar  ar  ...  ar and 0  br  br  ...  br 1 2 n 1 2 n if r  1. Hence by part (e) I haven n nar jj=1 j=1br jnj=1(ar br ). j jHence1 n r a n j=1 j1/r1 n r b n j=1 j1/r1 n (aj bj )r n j=11/rAlso, by the note of part (e), this inequality must be reversed of {aj } is increasing and {bj } is decreasing.5`

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