Please help, Prove by Arithmetic Mean and Geometric Mean

Question

For $a,b,c \in \mathbb{R}^+$, prove that $(a + b + c)^3 - 27 abc \ge 0$ . Can anyone please help? I really don't know how to solve this.

Answer 1

From the AM-GM inequality for n positive reals, we have $$\frac{\sum_{i=1}^n{x_i}}{n} \ge (\prod_n x_i)^{1/n}$$

Now, substituting $n = 3$ and $x_1 =a , x_2=b,x_3=c$, we get

$$\frac{(a+b+c)}{3} \ge (a.b.c)^{1/3} \implies (a+b+c)^3 \ge 27abc$$