[(b+c)/a] + [(c+a)/b] + [(a+b)/c]    
= [bc(b+c)/abc] + [ac(c+a)/abc] + [ab(a+b)/abc]
= (b²c+bc²+ac²+a²c+a²b+ab²)/(abc)
en permutant l'ordre des termes ->
= (b²c+a²c+bc²+a²b+ac²+ab²)/(abc)
= [c(a²+b²) + b(a²+c²) + a(b²+c²)]/abc
= [(a²+b²)/ab] + [(a²+c²)/ac] + [(b²+c²)/bc]  

[(b+c)/a] + [(c+a)/b] + [(a+b)/c]  = [(a²+b²)/ab] + [(a²+c²)/ac] + [(b²+c²)/bc]
  (1)


(x-y)² = x² + y² - 2xy
(x-y)²/(xy) = (x²+y²)/(xy) - 2
(x²+y²)/(xy) = (x-y)²/(xy) + 2

et si x > 0 et y > 0, on a alors (x-y)²/(xy) > 0
et donc:
(x²+y²)/(xy) > 2

et donc on a:
(a²+b²)/(ab) > 2
(a²+c²)/(ac) > 2
(b²+c²)/(bc) > 2

La somme -> [(a²+b²)/ab] + [(a²+c²)/ac] + [(b²+c²)/bc] > 6
remis dans (1) ->
[(b+c)/a] + [(c+a)/b] + [(a+b)/c] > 6
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Sauf distraction.