Tessellate S.T.E.M.S - Mathematics - School - Set 3 - Problem 2

Algebra Level 3

S = a ( a + b ) a 2 + ( a + b ) 2 + b ( b + c ) b 2 + ( b + c ) 2 + c ( c + d ) c 2 + ( c + d ) 2 + d ( d + a ) d 2 + ( d + a ) 2 S = \dfrac{a(a+b)}{a^2+(a+b)^2} + \dfrac{b(b+c)}{b^2+(b+c)^2}+ \dfrac{c(c+d)}{c^2+(c+d)^2}+ \dfrac{d(d+a)}{d^2+(d+a)^2} A = { 1 , 4 5 , 3 7 , 7 5 , 9 4 , 8 5 , 11 7 , 13 5 , 12 7 } A = \bigg\{1, \dfrac{4}{5}, \dfrac{3}{7}, \dfrac{7}{5}, \dfrac{9}{4}, \dfrac{8}{5}, \dfrac{11}{7}, \dfrac{13}{5}, \dfrac{12}{7} \bigg\} How many elements L A L \in A satisfy S L S \leq L for all positive reals a , b , c , d a,b,c,d ?


This problem is a part of Tessellate S.T.E.M.S.

6 4 5 3

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1 solution

Sutirtha Datta
Jan 10, 2018

WLOG we assume a > b > c > d a > b > c > d So by rearrangement inequality

a ( a + b ) a 2 + ( a + b ) 2 \sum\frac{a(a+b)}{a^2+(a+b)^2}

2 5 4 = 8 / 5 \leq\frac{2}{5}*4 =8/5

Then also, the answer should be 5 instead of 4

Aaghaz Mahajan - 3 years, 4 months ago

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Tell me the 5 candidates :)

Sutirtha Datta - 3 years, 4 months ago

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1, 4/5, 3/7, 7/5, and 8/5

Aaghaz Mahajan - 3 years, 4 months ago

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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

Sutirtha Datta - 3 years, 4 months ago

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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...

Aaghaz Mahajan - 3 years, 4 months ago

Can you explain how it is <= 2 by 5

Anu Radha - 3 years, 4 months ago

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