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All-Soviet-Union Mathematical Olympiad

Algebra Difficulty 4.8 AIME Prove it Soviet Union

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

Solution

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

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

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

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