Question

If a, b, c be positive numbers, (a + b + c)3 ≥ x abc, then find the greatest value of x.

Answer 1

answer : 27

explanation : it is based on concept of Arithmetic mean and geometric mean.

arithmetic mean of any positive numbers is always greater than geometric mean.

or, AM ≥ GM.

here, three positive numbers are a, b, and c.

arithmetic mean of a , b and c = (a + b + c)/3

and geometric mean of a, b and c = $\sqrt[3]{abc}$

now AM ≥ GM

or, (a + b + c)/3 ≥ $\sqrt[3]{abc}$

taking cube both sides ,

or, (a + b + c)³/3³ ≥ ([tex]\sqrt[3]{abc}[tex])³

or, (a + b + c)³ ≥ 3³ (abc)

or, (a + b + c)³ ≥ 27abc

on comparing (a + b + c)³ ≥ x abc , we get x = 27.

hence, greatest value of x = 27.

Answer 2

Answer:

answer : 27

explanation : it is based on concept of Arithmetic mean and geometric mean.

arithmetic mean of any positive numbers is always greater than geometric mean.

or, AM ≥ GM.

here, three positive numbers are a, b, and c.

arithmetic mean of a , b and c = (a + b + c)/3

and geometric mean of a, b and c =  

now AM ≥ GM

or, (a + b + c)/3 ≥  

taking cube both sides ,

or, (a + b + c)³/3³ ≥ ([tex]\sqrt[3]{abc}[tex])³

or, (a + b + c)³ ≥ 3³ (abc)

or, (a + b + c)³ ≥ 27abc

on comparing (a + b + c)³ ≥ x abc , we get x = 27.

hence, greatest value of x = 27.