q6)
a+b+c=(a+b-c)+(b+c-a)+(c+a-b)
using Am-GM WE HAVE
(a+b+c)/ 3 > or = third root of (a+b_c)(b+c-a)(c+A-b)
by cubing both sides we get the required result
: 1 (0 1 )
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