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

Question

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

I've tried using AM-GM by splitting the inequality into two inequalities, but I got stuck here. Any help would be appreciated.

I had an idea of removing x from all sides. Would that be helpful or no?

Thanks in advance.

Answer 1

The left inequality it's just Holder: $$(x+a)(x+b)(x+c)\geq\left(\sqrt[3]{x\cdot x\cdot x}+\sqrt[3]{abc}\right)^3=\left(x+\sqrt[3]{abc}\right)^3.$$ The right inequality is true by AM-GM: $$(x+a)(x+b)(x+c)\leq\left(\frac{x+a+x+b+x+c}{3}\right)^3=\left(x+\frac{a+b+c}{3}\right)^3.$$

Answer 2

Yes, your idea works!

The left inequality.

We need to prove that: $$ x^3+3\sqrt[3]{abc}x^2+3\sqrt[3]{a^2b^2c^2}x+abc\leq x^3+(a+b+c)x^2+(ab+ac+bc)x+abc,$$ which is true by AM-GM: $$a+b+c\geq3\sqrt[3]{abc}$$ and $$ab+ac+bc\geq3\sqrt[3]{a^2b^2c^2}.$$ The right inequality we can prove by the similar way.