<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] $$F=(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ca)x+abc~~~~(1)$$ By AM-GM 
$$F \ge x^3 + 3(abc)^{1/3} x^2+3(abc)^{2/3} x+ abc= (x+(abc)^{1/3})~~~~(2)$$
Using $$\frac{(a+b+c)^2}{3} \ge (ab+bc+ca)$$ and AM-GM  in (1), we get
$$F \le x^3+3 \frac{(a+b+c)}{3} x^2 +3\frac{(a+b+c)^2}{9}x+ \left(\frac{a+b+c}{3}\right)^3 = \left(x+\frac{a+b+c}{3}\right)^3$$
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