Question

If a, b, c positive number such that a + b + c = 4, then a³ + b³ + c³ is:
1. Less than of equal to \(\dfrac{16}{19}\)
2. Greater than or equal to \(\dfrac{16}{19}\)
3. Less than or equal to \(​​\dfrac{64}{9}\)
4. Greater than or equal to \(​​\dfrac{64}{9}\)

Answer 1

Correct Answer - Option 4 : Greater than or equal to \(​​\dfrac{64}{9}\)

Concept:

Arithmetic Mean ≥  Geometric Mean

AM ≥ GM

\(\frac{a~+~b~+~c}{3}≥ (abc)^\frac{1}{3}\)

(a3 + b3 + c3) = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc

Calculation: 

Given:

a + b + c = 4, a3 + b3 + c3 = ?

\(\frac{a~+~b~+~c}{3}≥ (abc)^\frac{1}{3}\)

\(abc \leq (\frac{4}{3})^3\leq \frac{64}{27}\)

\(3abc \leq (\frac{4}{3})^3\leq \frac{64}{9}\)----- (1)

(a3 + b3 + c3) = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc

For, a = b = c, (a + b + c)(a2 + b2 + c2 – ab – bc – ca) = 0

As a, b, c are positive numbers, 

The value of (a + b + c)(a2 + b2 + c2 – ab – bc – ca) > 0 ----(2)

∴ From (1) and (2)

(a3 + b3 + c3) = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) + 3abc

(a3 + b3 + c3) ≥ \(\frac{64}{9}\)