a > 0, b > 0, c > 0 だから相加平均と相乗平均の関係から

(a + b)/c + (b + c)/a + (c + a)/b
= \(\frac{a}{c}\) + \(\frac{b}{c}\) + \(\frac{b}{a}\) + \(\frac{c}{a}\) + \(\frac{c}{b}\) + \(\frac{a}{b}\)
= (\(\frac{a}{c}\) + \(\frac{c}{a}\)) + (\(\frac{b}{c}\) + \(\frac{c}{b}\)) + (\(\frac{b}{a}\) + \(\frac{a}{b}\))
≧2\(\sqrt{\quad}\)((\(\frac{a}{c}\))×(\(\frac{c}{a}\))) + 2\(\sqrt{\quad}\)((\(\frac{b}{c}\))×(\(\frac{c}{b}\))) + 2\(\sqrt{\quad}\)((\(\frac{b}{a}\))×(\(\frac{a}{b}\)))
= 2 + 2 + 2 = 6

であるから, 左辺が \(\frac{3}{2}\) になるなんて絶対あり得ない。