AM-GM inequality proof for nonnegative real numbers x, a, b, c

Question

Let $x,$ $a,$ $b,$ $c$ be nonnegative real numbers. Prove that $\left(x + \sqrt[3]{abc}\right)^3 \le (x + a)(x + b)(x + c) \le \left( x + \frac{a + b + c}{3} \right)^3.$

So far all i've done is expand out the inequality: $$x^3+3x^2\sqrt[3]{abc}+3x\sqrt[3]{a^2b^2c^2}+abc\le (x^3+cx^2+x^2(a+b)+cx(a+b)+abx+abc)\le x^3+x^2(a+b+c)+\frac{x(a+b+c)^2}{3}+\frac{(a+b+c)^3}{27}.$$ I think I need to use the RHS-AM-GM-HM inequality. How do I finish off this proof?

Answer 1

The left inequality follows from $$3\sqrt[3]{abc}\le a+b+c$$ and $$3\sqrt[3]{a^2b^2c^2}\le ab+bc+ca$$ both of which are AM/GM.

The right inequality follows from $$ab+bc+ca\le\frac{(a+b+c)^2}{3}$$ and $$abc\le\frac{(a+b+c)^3}{27}.$$ The latter is AM/GM and the former is equivalent to the well-known $$ab+bc+ca\le a^2+b^2+c^2.$$

Answer 2

The left inequality is true by Holder $$\prod_{cyc}(x+a)\geq\left(\sqrt[3]{x^3}+\sqrt[3]{abc}\right)^3=\left(x+\sqrt[3]{abc}\right)^3.$$ The right inequality.

Since $\ln$ is a concave function and $f(t)=e^t$ increases, by Jensen we obtain: $$\prod_{cyc}(x+a)=e^{\sum\limits_{cyc}\ln(x+a)}\leq e^{3\ln\frac{\sum\limits_{cyc}(x+a)}{3}}=\left(x+\frac{a+b+c}{3}\right)^3.$$