13√(x+a)(x+b)(x+c) <= x + (a+b+c)/3
=> 3√(x+a)(x+b)(x+c) - x <= (a+b+c)/3
at limitiing case, we take equality...
so, ans = b.
lim x→infinity ( 3√ (x+a) (x+b) (x+c)-x) =
a.√abc
b.(a+b+c)/3
c.abc
d.3√abc
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5 Answers
1take a=b=c=2 ..... 2 is the ans ... a&c eliminated ..
take a=b=0 , c =3 ..... 1 is the ans ... d eliminated ..
b ans ..
1[(x+a)+(x+b)+(x+c)]/3 ≥[(x+a)(x+b)(x+c)]1/3
or
[(x+a)+(x+b)+(x+c)]/3 -x ≥[(x+a)(x+b)(x+c)]1/3 -x
or
(a+b+c)/3≥[(x+a)(x+b)(x+c)]1/3 -x
Lim x-> inf expression=(a+b+c)/3
B)
cheers!!
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