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(1/a +1/b)(1/c +1/d)+1/ab +1/cd=6 /\sqrt{abcd}

Source: 2010 Moldova JBMO TST p1

February 25, 2021
algebra

Problem Statement

The positive real numbers a,b,c,da, b, c, d satisfy the equality (1a+1b)(1c+1d)+1ab+1cd=6abcd\left(\frac{1}{a}+ \frac{1}{b}\right) \left(\frac{1}{c}+ \frac{1}{d}\right) + \frac{1}{ab}+ \frac{1}{cd} = \frac{6}{\sqrt{abcd}} Find the value of the a2+ac+c2b2bd+d2\frac{a^2+ac+c^2}{b^2-bd+d^2}