Prove that $(x + \sqrt[3]{abc})^3 \le (x + a)(x + b)(x + c) \le ( x + \frac{a + b + c}{3})^3$

Question

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.

Answer 1

Hint for the LHS:

$$3 = \sum_{a,b,c}\dfrac{x}{x+a}+\sum_{a,b,c}\dfrac{a}{x+a}\geq\text{Do AM-GM here}.$$

RHS is more trivial than this as it is just exactly one application of AM-GM in disguise.

Answer 2

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) $$(x+a)(x+b)(x+c)=x^3\left(1+\frac{a}{x}\right)\left(1+\frac{b}{x}\right)\left(1+\frac{c}{x}\right)\geq\\ x^3\left(1+\sqrt[3]{\frac{a}{x}\frac{b}{x}\frac{c}{x}}\right)^3= \left(x+\sqrt[3]{abc}\right)^3$$

Answer 3

$$F=(x+a)(x+b)(x+c)=x^3+(a+b+c)x^2+(ab+bc+ca)x+abc~~~~(1)$$ 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 $$F \le x^3+3 \frac{(a+b+c)}{3} x^2 +3\frac{(a+b+c)^2}{9}x+ \left(\frac{a+b+c}{3}\right)^3 = \left(x+\frac{a+b+c}{3}\right)^3$$