nice-calculus-a-b-c-d-N-and-1-a-1-b-1-c-1-d-1-2-find-

Question Number 115058 by mnjuly1970 last updated on 23/Sep/20

\dots . n i c e c a l c u l u s \dots

a, b, c, d \in N a n d

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = \frac{1}{2}

f i n d

m a x (a + b + c + d) = ? ? ?

\dots m . n . j u l y . 1970 \dots

Answered by 1549442205PVT last updated on 25/Sep/20

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = \frac{1}{2} (1)

Since a, b, c, d are equal in role WLOG

we can suppose that a ⩾ b ⩾ c ⩾ d

\Rightarrow \frac{4}{a} ⩽ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = \frac{1}{2} ⩽ \frac{4}{d}

\Rightarrow a ⩾ 8 ⩾ d ⩾ 3

i) Case d = 3 \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}

\Rightarrow \frac{3}{a} ⩽ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{6} ⩽ \frac{3}{c}

\Rightarrow a ⩾ 18 ⩾ c ⩾ 7

a) For c = 7 \Rightarrow \frac{1}{a} + \frac{1}{b} = \frac{1}{6} - \frac{1}{7} = \frac{1}{42}

\Rightarrow \frac{2}{a} ⩽ \frac{1}{a} + \frac{1}{b} = \frac{1}{42} ⩽ \frac{2}{b}

\Rightarrow a ⩾ 84 ⩾ b ⩾ 43

∙ if b = 43 then \frac{1}{a} = \frac{1}{42} - \frac{1}{43} = \frac{1}{1806}

\Rightarrow a = 1806

Thus, a + b + c + d = 1806 + 43 + 7 + 3 = 1859

∙ If b = 44 \Rightarrow \frac{1}{a} = \frac{1}{42} - \frac{1}{44} = \frac{1}{924} \Rightarrow a = 924

42 (a + b) = ab \Rightarrow (a - 42) (b - 42) = 1764

= 4 . 21^{2} = 2^{2} . 3^{2} . 7^{2}

b) For c = 8 \Rightarrow \frac{1}{a} + \frac{1}{b} = \frac{1}{6} - \frac{1}{8} = \frac{1}{24}

\Rightarrow ab = 24 (a + b) \Leftrightarrow (a - 24) (b - 24) = 576

= 24^{2} = 2^{6} . 3^{2}

b = 25 \Rightarrow a = 600 \Rightarrow a + b + c + d = 636 < 1859

c) For c = 9, 10, \dots, 18 by similar argument

we see that a + b + c + d < 1859

ii) d = 4 \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

\Rightarrow \frac{3}{a} ⩽ \frac{1}{4} ⩽ \frac{3}{c} \Rightarrow a ⩾ 12 ⩾ c ⩾ 5

a) For c = 5 \Rightarrow \frac{1}{a} + \frac{1}{b} = \frac{1}{4} - \frac{1}{5} = \frac{1}{20}

\Rightarrow ab = 20 (a + b) \Leftrightarrow (a - 20) (b - 20) = 400 = 2^{4} . 5^{2}

∙ if b = 21 \Rightarrow a = 420 \Rightarrow a + b + c + d = 450 < 1859

b) For c = \overset{―}{6 \dots 12}, by similar argument

we get a + b + c + d < 1859

ii) Cases d = \overset{―}{5 \dots 8}, repeating above argument

we obtain a + b + c + d < 1859

Thus, if \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} = \frac{1}{2}

with a, b, c, d \in N then

\max (a + b + c + d) = 1859

Commented by mnjuly1970 last updated on 23/Sep/20

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