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Problem of the week no6

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My approach to the Purdue's sixth problem of the week.

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Problem of the week no6

  1. 1. Problem of the Week No.6 Eduardo Enrique Escamilla SaldaΓ±a March 2 2014. Monterrey, Nuevo LeΓ³n, MΓ©xico Let π‘Ž1 , π‘Ž2 , … , π‘Ž 𝑛 be positive numbers. Find the smallest possible value of 𝑛 βˆ‘ π‘˜=1 π‘Žπ‘˜ π‘Žπ‘– π‘˜ Where 𝑖1 , 𝑖2 , … , 𝑖 𝑛 is a permutation of 1,2, … , 𝑛. Lemma: For all π‘Ž 𝑖 > 0 π‘Žπ‘›π‘‘ 𝑖 β‰  𝑗: π‘Žπ‘— π‘Žπ‘– + β‰₯2 π‘Žπ‘— π‘Žπ‘– Proof: 2 (π‘Ž 𝑖 βˆ’ π‘Ž 𝑗 ) β‰₯ 0 ∴ π‘Ž 𝑖 2 βˆ’ 2π‘Ž 𝑖 π‘Ž 𝑗 + π‘Ž 2 β‰₯ 0 𝑖 ∴ π‘Ž 𝑖 2 + π‘Ž 2 β‰₯ 2π‘Ž 𝑖 π‘Ž 𝑗 𝑖 Dividing both sides by π‘Ž 𝑖 π‘Ž 𝑗 > 0 yields: π‘Žπ‘— π‘Ž 𝑖 2 + π‘Ž2 π‘Žπ‘– 𝑖 = + β‰₯2 π‘Ž 𝑖 π‘Žπ‘— π‘Žπ‘— π‘Žπ‘– As desired. Without loss of generalization let π‘Ž1 ≀ π‘Ž2 ≀ β‹― ≀ π‘Ž 𝑛 Equivalently: 1 1 1 ≀ ≀⋯≀ π‘Žπ‘› π‘Ž π‘›βˆ’1 π‘Ž1 Then 𝑛 βˆ‘ π‘˜=1 π‘Žπ‘˜ π‘Žπ‘– π‘˜ Is minimal when every term in the two inequalities are multiplied pairwise in the same order i.e.
  2. 2. 𝐴 = π‘Ž1 βˆ— 1 1 1 + π‘Ž2 βˆ— + β‹―+ π‘Ž 𝑛 βˆ— π‘Žπ‘› π‘Ž π‘›βˆ’1 π‘Ž1 For if one changes any two terms in the arrangement say, π‘Ž 𝑝 ≀ π‘Ž π‘ž then: If 𝐴 = π‘Ž1 βˆ— 1 1 1 1 + β‹―+ π‘Ž 𝑝 βˆ— +β‹―+ π‘Žπ‘ž βˆ— + β‹―+ π‘Ž 𝑛 βˆ— π‘Žπ‘› π‘Ž π‘›βˆ’π‘+1 π‘Ž π‘›βˆ’π‘ž+1 π‘Ž1 𝐡 = π‘Ž1 βˆ— 1 1 1 1 +β‹―+ π‘Žπ‘ž βˆ— + β‹―+ π‘Ž 𝑝 βˆ— + β‹―+ π‘Ž 𝑛 βˆ— π‘Žπ‘› π‘Ž π‘›βˆ’π‘+1 π‘Ž π‘›βˆ’π‘ž+1 π‘Ž1 Permuting π‘Ž 𝑝 , π‘Ž π‘ž gives But π΅βˆ’ 𝐴 = π‘Žπ‘žβˆ— 1 1 1 1 1 1 + π‘Žπ‘βˆ— βˆ’ π‘Žπ‘βˆ— βˆ’ π‘Žπ‘žβˆ— = (π‘Ž 𝑝 βˆ’ π‘Ž π‘ž )( βˆ’ )β‰₯0 π‘Ž π‘›βˆ’π‘+1 π‘Ž π‘›βˆ’π‘ž+1 π‘Ž π‘›βˆ’π‘+1 π‘Ž π‘›βˆ’π‘ž+1 π‘Ž π‘›βˆ’π‘ž+1 π‘Ž π‘›βˆ’π‘+1 So the sum 𝐴 is minimal since its status as the smallest value is invariant under any perturbation in the arrangement of its terms. Therefore the minimal sum is 𝐴 = π‘Ž1 βˆ— (π‘Ž1 βˆ— 1 1 1 + π‘Ž2 βˆ— + β‹―+ π‘Ž 𝑛 βˆ— π‘Žπ‘› π‘Ž π‘›βˆ’1 π‘Ž1 1 1 1 1 1 1 1 + π‘Ž 𝑛 βˆ— ) + (π‘Ž2 βˆ— + π‘Ž π‘›βˆ’1 βˆ— ) + β‹― + (π‘Ž 𝑖 βˆ— + π‘Ž π‘›βˆ’π‘–+1 βˆ— ) + β‹― + π‘Ž 𝑛+1 βˆ— , π‘Žπ‘› π‘Ž1 π‘Ž π‘›βˆ’1 π‘Ž2 π‘Ž π‘›βˆ’π‘–+1 π‘Žπ‘– π‘Ž 𝑛+1 2 𝑛 π‘œπ‘‘π‘‘ 2 = (π‘Ž1 βˆ— { 1 1 1 1 1 1 1 1 + π‘Ž 𝑛 βˆ— ) + (π‘Ž2 βˆ— + π‘Ž π‘›βˆ’1 βˆ— ) + β‹― + (π‘Ž 𝑖 βˆ— + π‘Ž π‘›βˆ’π‘–+1 βˆ— ) + β‹― + (π‘Ž 𝑛 βˆ— + π‘Ž 𝑛+1 βˆ— ), 𝑒𝑣𝑒𝑛 π‘Žπ‘› π‘Ž1 π‘Ž π‘›βˆ’1 π‘Ž2 π‘Ž π‘›βˆ’π‘–+1 π‘Žπ‘– π‘Ž 𝑛+1 π‘Žπ‘› 2 2 2 Applying the lemma we have that for 𝑛 π‘œπ‘‘π‘‘ there are exactly plus the constant π‘Ž 𝑛+1 βˆ— 2 1 π‘Ž 𝑛+1 π‘›βˆ’1 terms with the property that 2 π‘Žπ‘— π‘Žπ‘– + β‰₯2 π‘Žπ‘— π‘Žπ‘– = 1 implying that for this case 2 𝐴β‰₯2βˆ— π‘›βˆ’1 +1= 𝑛 2 𝑛 Similarly for 𝑛 𝑒𝑣𝑒𝑛 there are exactly terms with the property that 2 π‘Žπ‘— π‘Žπ‘– + β‰₯2 π‘Žπ‘— π‘Žπ‘– So 𝐴β‰₯2βˆ— 𝑛 = 𝑛 2 Therefore the smallest possible value for 𝑛 βˆ‘ π‘˜=1 π‘Žπ‘˜ π‘Žπ‘– π‘˜ 2
  3. 3. Where 𝑖1 , 𝑖2 , … , 𝑖 𝑛 is a permutation of 1,2, … , 𝑛 is 𝑛 Provided that all the terms are positive.

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