Question

If a, b, c > 0 prove that a^3 + b^3 + c^3 > 3abc .

Answer 1

Answer:

Step-by-step explanation:

If a , b , c > 0, then sum of a , b , c will be greater than zero.

=> a + b + c > 0

=> a + b > - c

Cubing on both sides

=> (a + b)³ > - c³

=> a³ + b³ + 3ab(a + b) > - c³

=> a³ + b³ + c³ > - 3ab(a+b)

=> a³ + b³ + c³ > - 3ab(-c)

=> a³ + b³ + c³  > 3abc.

Hence proved.