Question regarding inequality: if a+b+c=1 then a^3+b^3+c^3+6abc >= 1/4

[HelloMath]

Let a, b, and c be positive real numbers such that a + b + c = 1. Prove the following inequality:

a³ + b³ + c³ + 6abc > 1/4

 

[stapel]

Let a, b, and c be positive real numbers such that a + b + c = 1. Prove the following inequality:

a³ + b³ + c³ + 6abc > 1/4

Since a + b + c = 1, then you can multiply the inequality through by 4 and then substitute:

. . . . .\(\displaystyle a^3\, +\, b^3\, +\, c^3\, +\, 6abc\, \geq\, \dfrac{1}{4}\)

. . . . .\(\displaystyle 4\, (a^3\, +\, b^3\, +\, c^3)\, +\, 24abc\, \geq\, 1\)

. . . . .\(\displaystyle 4\, (a^3\, +\, b^3\, +\, c^3)\, +\, 24abc\, \geq\, a\, +\,b\, +\, c\)

Since the right-hand side is just 1, and since you're working with cubes, why not cube the 1?

. . . . .\(\displaystyle 4\, (a^3\, +\, b^3\, +\, c^3)\, +\, 24abc\, \geq\, (a\, +\,b\, +\, c)^3\)

Where does this lead?