Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$

Question

I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can be found in that pdf but I will write here to be more explicitly.

Exercise 1.3.4(a) Let $a,b,c$ be positive real numbers. Prove that $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4.$$

(b) For real numbers $a,b,c \gt0$ and $n \leq3$ prove that: $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$

Answer 1

Write $$\frac ab+\frac ab+\frac bc\geq \frac{3a}{\sqrt[3]{abc}}$$ by AM-GM.

You get $$\operatorname{LHS} \geq \frac{a+b+c}{\sqrt[3]{abc}}+n\left(\frac{\sqrt[3]{abc}}{a+b+c}\right).$$

Set $$z:=\frac{a+b+c}{\sqrt[3]{abc}}$$ and then notice that for $n\leq 3$, $$z+\frac{3n}{z}\geq 3+n.$$ Indeed the minimum is reached for $z=\sqrt{3n}\leq 3$; since $z\geq 3$, the minimum is reached in fact for $z=3$.

Answer 2

Following above motivations and applying AM-GM three times: \begin{align} &\frac13\left(\frac ab+\frac ab+\frac bc\right)+\frac13\left(\frac bc+\frac bc+\frac ca\right)+\frac13\left(\frac ca+\frac ca+\frac ab\right)+\frac{3\sqrt[3]{abc}}{a+b+c}\\ &\ge \frac{a}{\sqrt[3]{abc}}+\frac{b}{\sqrt[3]{abc}}+\frac{c}{\sqrt[3]{abc}}+\frac{3\sqrt[3]{abc}}{a+b+c}\\ &=\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{3\sqrt[3]{abc}}{a+b+c}\\ &\ge4\left(\left(\frac{a+b+c}{3\sqrt[3]{abc}}\right)^2\right)^\frac14\\ &\ge4. \end{align}