Question

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 1

(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.