If a,b,c are non zero real numbers, then minimum value of the expression((a4+a2+1)(b4+7b2+1)(c4+11c2+1)(a2b2c2)) is
A
315
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B
351
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C
415
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D
451
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Solution
The correct option is B351 ((a4+a2+1)(b4+7b2+1)(c4+11c2+1)(a2b2c2))=(a2+1a2+1)+(b2+1b2+7)+(c2+1c2+11) Using A.M≥G.M we know that x2+1x2≥2 The minimum value of the expression will occur when all the terms are mininum, =(2+1)(2+7)(2+11)=351