For three distinct positive real numbers a, b, c (1 + a3) (1 + b3) (1

1.	For three distinct positive real numbers a, b, c (1 + a3) (1 + b3) (1 + c3) is greater than(a) abc (b) (1 + abc) (c) (1 + abc)3 (d) (1 + abc)2
Answer» (c) (1 + abc)³ (1 + a³) (1 + b³) (1 + c³) = 1 + a³ + b³ + c³ + a³b³ + b³c³ + c³a³ + a³b³c³ …(i) Now for distinct positive reals, a, b, c, AM > GM \(\frac{a^3+b^3+c^3}{3}>(a^3b^3c^3)^{\frac13}\) ⇒ a³ + b³ + c³ > 3abc …(ii) (∵ a, b, c > 0 ⇒ a³, b³, c³ > 0) Also \(\frac{a^3b^3+b^3c^3+c^3a^3}{3}>(a^3b^3.b^3c^3.c^3a^3)^{\frac13}\) ⇒ a³b³ + b³c³ + c³a³ > 3a²b²c² …(iii) ∴ Putting the values from (ii) and (iii) on the RHS of (i), we have (1 + a³) (1 + b³) (1 + c³) > 1 + 3abc + 3a²b²c² + a³b³c³ = (1 + abc)³ ∴ (1 + a³) (1 + b³) (1 + c³) > (1 + abc)³.

For three distinct positive real numbers a, b, c (1 + a3) (1 + b3) (1 + c3) is greater than(a) abc (b) (1 + abc) (c) (1 + abc)3 (d) (1 + abc)2

Answer»

(c) (1 + abc)³

(1 + a³) (1 + b³) (1 + c³) = 1 + a³ + b³ + c³ + a³b³ + b³c³ + c³a³ + a³b³c³ …(i)

Now for distinct positive reals, a, b, c, AM > GM

\(\frac{a^3+b^3+c^3}{3}>(a^3b^3c^3)^{\frac13}\)

⇒ a³ + b³ + c³ > 3abc …(ii)

(∵ a, b, c > 0 ⇒ a³, b³, c³ > 0)

Also \(\frac{a^3b^3+b^3c^3+c^3a^3}{3}>(a^3b^3.b^3c^3.c^3a^3)^{\frac13}\)

⇒ a³b³ + b³c³ + c³a³ > 3a²b²c² …(iii)

∴ Putting the values from (ii) and (iii) on the RHS of (i), we have

(1 + a³) (1 + b³) (1 + c³) > 1 + 3abc + 3a²b²c² + a³b³c³ = (1 + abc)³

∴ (1 + a³) (1 + b³) (1 + c³) > (1 + abc)³.