How to prove $\frac{a}{b} + \frac{b}{c} + \frac{c}{a}+ \frac{3\sqrt[3]{abc}}{a+b+c}\geq 4 $ using Middle School methods?

Question

How to prove this:

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

(Sorry I've asked this question on phone.)

I can understand the C-S step, but I've never learnt about the Schur inequality... I have googled it but I have no idea how to apply it here :/

I've done it by using harmonic mean to the Schur

Answer 1

The first step:

By C-S $$\sum_{cyc}\frac{a}{b}=\sum_{cyc}\frac{a^2}{ab}\geq\frac{(a+b+c)^2}{ab+ac+bc}.$$ Thus, it remains to prove that $$\frac{(a+b+c)^2}{ab+ac+bc}+\frac{3\sqrt[3]{abc}}{a+b+c}\geq4,$$ which is true and the proof for you.