Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Prove it Soviet Union

Problem:
Given positive numbers aa, bb, cc, dd prove that at least one of the inequalities does not hold:
a+b<c+da + b < c + d ;
(a+b)(c+d)<ab+cd(a + b)(c + d) < ab + cd ;
(a+b)cd<ab(c+d)(a + b)cd < ab(c + d) .

Solution

Solution:
From the first and second inequalities we have ab+cd>a(c+d)+b(a+b)ab + cd > a(c + d) + b(a + b), so cd>adcd > ad, and hence c>ac > a.

We also have ab+cd>a(a+b)+b(c+d)ab + cd > a(a + b) + b(c + d), so cd>bccd > bc, and hence d>bd > b.

So 1/a+1/b>1/c+1/d1/a + 1/b > 1/c + 1/d, which contradicts the third inequality.

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