<s>[INST] Prove that $(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le ( x + \frac{a + b + c}{3})^3$ Let $x,$ $a,$ $b,$ $c$ be nonnegative real numbers. Prove that
$$(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le \left( x + \frac{a + b + c}{3} \right)^3.$$
I know that this problem is a RHS-AM-GM-HM problem, but I am unsure how to solve it. I think that the leftmost part is AM, but that is as far as I was able to get. I also think that taking the cube root of everything might change it into something more manageable, but I am still unsure about that. 
Can anyone help me with this or give me tips about how to solve it? Thank you.
 [/INST] Using AM
$$\left( x + \frac{a + b + c}{3} \right)^3=
\left(\frac{(x+a) + (x+b) + (x+c)}{3} \right)^3\overset{AM}{\geq} \\
\left(\sqrt[3]{(x+a)(x+b)(x+c)}\right)^3=
(x+a)(x+b)(x+c)$$
Using HUYGEN’S INEQUALITY (which is easy to prove using AM-GM)
$$(x+a)(x+b)(x+c)=x^3\left(1+\frac{a}{x}\right)\left(1+\frac{b}{x}\right)\left(1+\frac{c}{x}\right)\geq\\
x^3\left(1+\sqrt[3]{\frac{a}{x}\frac{b}{x}\frac{c}{x}}\right)^3=
\left(x+\sqrt[3]{abc}\right)^3$$
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