if a,b,c are real no then ∆=
|a-1,a,a+1,b-1,b,,b+1,c-1,c,c+1|
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Answer
[(1+a)(1+b)(1+c)]
7
>7
7
a
4
b
4
c
4
i.e.,(1+∑a+∑ab+abc)>7(abc)
4/7
Since, AM>GM
a+b+c+≥3(abc)
1/3
ab+bc+ac≥3(a
2
b
2
c
2
)
1/3
Hence,
1+∑a+∑ab+abc≥1+3(abc)
1/3
+3(abc)
2/3
+abc1+∑a+∑ab+abc≥1+(abc)
1/3
+(abc)
1/3
+(abc)
1/3
+(abc)
2/3
+(abc)
2/3
+(abc)
2/3
+abc
So,
LHS≥7[1(abc)
1/3
(abc)
1/3
(abc)
1/3
(abc)
2/3
(abc)
2/3
(abc)
2/3
(abc)]
1/7
≥7[(abc)
9/3
(abc)]
1/7
≥7(abc)
4/7
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