S = a 2 + ( a + b ) 2 a ( a + b ) + b 2 + ( b + c ) 2 b ( b + c ) + c 2 + ( c + d ) 2 c ( c + d ) + d 2 + ( d + a ) 2 d ( d + a ) A = { 1 , 5 4 , 7 3 , 5 7 , 4 9 , 5 8 , 7 1 1 , 5 1 3 , 7 1 2 } How many elements L ∈ A satisfy S ≤ L for all positive reals a , b , c , d ?
This problem is a part of Tessellate S.T.E.M.S.
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Then also, the answer should be 5 instead of 4
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Tell me the 5 candidates :)
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1, 4/5, 3/7, 7/5, and 8/5
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@Aaghaz Mahajan – Read the ques carefully ; it is about finding lower bound of L, so consider elements greater than 8/5
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@Sutirtha Datta – Oh shucks!! You are right.... IDK why I did that even after solving the question I even reported the problem.....My bad...
Can you explain how it is <= 2 by 5
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WLOG we assume a > b > c > d So by rearrangement inequality
∑ a 2 + ( a + b ) 2 a ( a + b )
≤ 5 2 ∗ 4 = 8 / 5