AM≥GM [3 x 1/3 +4 x a/4]/7 ≥ [(1/3)3(a/4)4]1/7
(1+a)7≥ 77a4/33.44
therefore (1+a)7(1+b)7(1+c)7≥[721.a4b4c4]/39.412 which is always greater than 77a4b4c4.
since [721.a4b4c4]/39.412>77a4b4c4.
714>39412
(72/3.412)7 > 32/42 which is is always true